Linear Motion
with constant acceleration
Physics 6A
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As long as its acceleration is constant*, the motion of any
object can be described by the following simple formulas:
(1)
x  x 0  v 0 t  21 at 2
(2)
v  v 0  at
(3)
v 2  v 02  2ax  x 0 
When using these formulas it is important to be clear about your variables.
Especially important is where and when the position and time are zero.
A couple of examples will help clarify this.
*both the magnitude and direction must be constant
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Sample Problem #1
A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.
Find the following:
a)
the average acceleration while it is speeding up
b)
the distance traveled while speeding up
c)
the average acceleration while braking
d)
the time elapsed while braking
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Sample Problem #1
A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.
Find the following:
a)
the average acceleration while it is speeding up
b)
the distance traveled while speeding up
c)
the average acceleration while braking
d)
the time elapsed while braking
By definition, average acceleration is just (change in velocity)/(change in time).
Since the car sped up from 0m/s to 30m/s in 5.6 seconds, this is a simple calculation:
a
30 ms
5.6s
 5.4 m
s
2
This is the answer for part a)
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Sample Problem #1
A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.
Find the following:
a)
the average acceleration while it is speeding up
b)
the distance traveled while speeding up
c)
the average acceleration while braking
d)
the time elapsed while braking
Another way to do part a) is to use formula (2) from before:
v  v 0  at
In this case we are starting from rest, so v0 = 0m/s,
and our final speed is v = 30m/s.
The elapsed time is t = 5.6s
Putting these numbers into the equation we get:
30m/s = 0 + a(5.6s)
Solving for a gives us the answer:
a ≈ 5.4m/s2
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Sample Problem #1
A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.
Find the following:
a)
the average acceleration while it is speeding up
b)
the distance traveled while speeding up
c)
the average acceleration while braking
d)
the time elapsed while braking
Now that we have the acceleration from part a), we can get the answer to b)
Which of our 3 equations is appropriate here?
(1)
x  x 0  v 0 t  21 at 2
(2)
v  v 0  at
(3)
v 2  v 02  2ax  x 0 
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Sample Problem #1
A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.
Find the following:
a)
the average acceleration while it is speeding up
b)
the distance traveled while speeding up
c)
the average acceleration while braking
d)
the time elapsed while braking
Now that we have the acceleration from part a), we can get the answer to b)
Which of our 3 equations is appropriate here?
Equation (1) or (3) will work in this case.
(1)
x  x 0  v 0 t  21 at 2
(2)
v  v 0  at
(3)
v 2  v 02  2ax  x 0 
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Sample Problem #1
A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.
Find the following:
a)
the average acceleration while it is speeding up
b)
the distance traveled while speeding up
c)
the average acceleration while braking
d)
the time elapsed while braking
Using Equation (1):
x  x 0  v 0 t  21 at 2


2

x  0  0  21 5.4 m
5
.
6
s
s
2
x  85m
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Sample Problem #1
A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.
Find the following:
a)
the average acceleration while it is speeding up
b)
the distance traveled while speeding up
c)
the average acceleration while braking
d)
the time elapsed while braking
Using Equation (3):
v 2  v 02  2ax  x 0 
30 ms 2  0  25.4 ms x  0
x
900 ms  10.8 m
s
2
2
2
2
x  83m
Note: The discrepancy is due to round-off error from the original calculation of a
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Sample Problem #1
A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.
Find the following:
a)
the average acceleration while it is speeding up
b)
the distance traveled while speeding up
c)
the average acceleration while braking
d)
the time elapsed while braking
Now we move on to the braking part of the problem:
We can organize our information in a table if we want:
x0 =
x=
v0 =
v=
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Sample Problem #1
A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.
Find the following:
a)
the average acceleration while it is speeding up
b)
the distance traveled while speeding up
c)
the average acceleration while braking
d)
the time elapsed while braking
Now we move on to the braking part of the problem:
We can organize our information in a table if we want:
x0 =0
x =50m
v0 =30m/s
v= 0
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Sample Problem #1
A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.
Find the following:
a)
the average acceleration while it is speeding up
b)
the distance traveled while speeding up
c)
the average acceleration while braking
d)
the time elapsed while braking
Now we move on to the braking part of the problem:
We can organize our information in a table if we want:
x0 =0
x =50m
v0 =30m/s
v= 0
Compare this information with our formulas from earlier.
Which formula is the easiest one to use to find acceleration?
(1)
x  x 0  v 0 t  21 at 2
(2)
v  v 0  at
(3)
v 2  v 02  2ax  x 0 
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Sample Problem #1
A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.
Find the following:
a)
the average acceleration while it is speeding up
b)
the distance traveled while speeding up
c)
the average acceleration while braking
d)
the time elapsed while braking
Now we move on to the braking part of the problem:
We can organize our information in a table if we want:
x0 =0
x =50m
v0 =30m/s
v= 0
Compare this information with our formulas from earlier.
Formula (3) gives us what we need.
(1)
x  x 0  v 0 t  21 at 2
(2)
v  v 0  at
(3)
v 2  v 02  2ax  x 0 
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Sample Problem #1
A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.
Find the following:
a)
the average acceleration while it is speeding up
b)
the distance traveled while speeding up
c)
the average acceleration while braking
d)
the time elapsed while braking
Here’s the calculation:
v 2  v 02  2ax  x 0 

0  30 ms
2  2a50m  0
a  9 m
s
2
Why did this come out negative?
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Sample Problem #1
A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.
Find the following:
a)
the average acceleration while it is speeding up
b)
the distance traveled while speeding up
c)
the average acceleration while braking
d)
the time elapsed while braking
Now that we have acceleration, we can find the time for part d)
Which formula should we use?
(1)
x  x 0  v 0 t  21 at 2
(2)
v  v 0  at
(3)
v 2  v 02  2ax  x 0 
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Sample Problem #1
A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.
Find the following:
a)
the average acceleration while it is speeding up
b)
the distance traveled while speeding up
c)
the average acceleration while braking
d)
the time elapsed while braking
Now that we have acceleration, we can find the time for part d)
Which formula should we use?
Either of (1) or (2) will work, so choose the easier one: Formula (2)
(1)
x  x 0  v 0 t  21 at 2
(2)
v  v 0  at
(3)
v 2  v 02  2ax  x 0 
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Sample Problem #1
A 2009 Audi A4 accelerates from rest to 30 m/s in about 5.6 seconds. From that speed, it stops in a distance of 50 meters.
Find the following:
a)
the average acceleration while it is speeding up
b)
the distance traveled while speeding up
c)
the average acceleration while braking
d)
the time elapsed while braking
Here is the calculation:
v  v 0  at
0  30 ms  ( 9 m
)t
s
2
t
30
s  3 .3 s
9
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
First we need to set up a coordinate system. A
convenient way to do it is to let the lowest point be
y=0, then call the upward direction positive.
With this choice, our initial values are:
y0 = 2.1m
v0 = 30m/s
v0 = 30 m/s
y=2.1m
y=0
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
For vertical motion on the surface of Earth, the
acceleration will always be downward, with
magnitude g = 9.8m/s2
Important: when you see g in a formula, it will be a
positive number. Every time. No exceptions.
v0 = 30 m/s
y=2.1m
y=0
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
Now we can think about the problem:
What is special about the maximum height?
In other words, do we know anything specific about
the ball when it reaches the maximum height?
v0 = 30 m/s
y=2.1m
y=0
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
Now we can think about the problem:
What is special about the maximum height?
In other words, do we know anything specific about
the ball when it reaches the maximum height?
YES! This is the most important piece of
information in this type of problem:
v0 = 30 m/s
y=2.1m
the vertical velocity is zero at the top.
y=0
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
v = 0 here
ymax=?
Now we can use a formula to find the max height.
Here they are again, adjusted for vertical motion:
v0 = 30 m/s
(1)
y  y 0  v 0 t  21 gt2
(2)
v  v 0  gt
(3)
v 2  v 02  2gy  y 0 
Which one matches our information?
y=2.1m
y=0
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
v = 0 here
ymax=?
Now we can use a formula to find the max height.
Here they are again, adjusted for vertical motion:
v0 = 30 m/s
(1)
y  y 0  v 0 t  21 gt2
(2)
v  v 0  gt
(3)
v 2  v 02  2gy  y 0 
Which one matches our information?
y=2.1m
We don’t have information about the time, so
we use Equation (3)
y=0
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
Here’s the calculation:
v = 0 here
ymax=?
v 2  v 02  2gy  y0 

0  30 ms
2  29.8 ms ymax  2.1m
2
ymax  48m
v0 = 30 m/s
y=2.1m
y=0
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
v = 0 here
ymax=48m
We can use (2) to find the time to get to the top
(we could have done this before part a) if we had wanted)
v  v 0  gt

 

0  30 ms  9.8 m
t
s
2
t  3.1s
v0 = 30 m/s
y=2.1m
y=0
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
v = 0 here
ymax=48m
Where is the ball when it is traveling at its
maximum speed?
v0 = 30 m/s
y=2.1m
y=0
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
v = 0 here
ymax=48m
Where is the ball when it is traveling at its
maximum speed?
Just before it hits the ground. A good way to think
about this is to realize that it is speeding up as it
falls, so the farther it falls, the faster it is moving.*
v0 = 30 m/s
y=2.1m
Fastest here
Note: as the ball passes by the starting point it
has a velocity of 30m/s downward. This is just the
opposite of the initial velocity! In fact, at any height,
the ball will be moving at the same speed on the
way up or the way down.
*We are ignoring air resistance in this problem, so there is no terminal speed.
y=0
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
v = 0 here
ymax=48m
We can use formula (3) to find the speed just as
the ball hits the ground. Our final position will be
y=0
We can use anywhere for the starting point, as long
as we put in the appropriate initial velocity.
v0 = 30 m/s
y=2.1m
y=0
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
v = 0 here
ymax=48m
Here is the answer in 2 ways:
With y0=2.1m:
v 2  v 02  2gy  y 0 

v 2  30 ms
2  29.8 ms 0  2.1m
2
v  30.7 ms
v0 = 30 m/s
With y0=48m:
y=2.1m
v 2  v 02  2gy  y0 


0  48m
v 2  0  2 9.8 m
s
y=0
2
v  30.7 ms
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
v = 0 here
ymax=48m
Time for the final question. Which formula should
we use to get the time when the ball hits?
v0 = 30 m/s
y=2.1m
y=0
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
v = 0 here
ymax=48m
Time for the final question. Which formula should
we use to get the time when the ball hits?
We can actually use either (1) or (2), but we have
to be careful!
Let’s try it first with Equation (1)…
v0 = 30 m/s
y=2.1m
y=0
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
We need to put in y,y0,v0 and g
We know it lands on the ground, so y = 0
If we use y0 = 2.1m and v0 = 30m/s:
y  y 0  v 0 t  21 gt2




2
0  2.1m  30 ms t  21 9.8 m
t
s
v0 = 30 m/s
2
How do we solve this???
y=2.1m
y=0
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
We need to put in y,y0,v0 and g
We know it lands on the ground, so y = 0
If we use y0 = 2.1m and v0 = 30m/s:
y  y 0  v 0 t  21 gt2




2
0  2.1m  30 ms t  21 9.8 m
t
s
v0 = 30 m/s
y=2.1m
2
If we want to solve this one, we need to use the
quadratic equation. This will work just fine, but it is
more work than we should really need to do here.
We get 2 answers:
y=0
t = 6.2s and t = -0.07s
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
We need to put in y,y0,v0 and g
We know it lands on the ground, so y = 0
If we use y0 = 2.1m and v0 = 30m/s:
y  y 0  v 0 t  21 gt2




2
0  2.1m  30 ms t  21 9.8 m
t
s
v0 = 30 m/s
y=2.1m
2
If we want to solve this one, we need to use the
quadratic equation. This will work just fine, but it is
more work than we should really need to do here.
We get 2 answers:
y=0
t = 6.2s and t = -0.07s
Prepared by Vince Zaccone
Choose the positive answer in
this case.
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
v = 0 here
ymax=48m
That took some tedious algebra to find the answer.
How can we avoid this?
v0 = 30 m/s
y=2.1m
y=0
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Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
v = 0 here
ymax=48m
That took some tedious algebra to find the answer.
How can we avoid this?
We can try using a different starting point.
How about starting at the top?
Putting in y0 = 48m and v0 = 0:
v0 = 30 m/s
y=2.1m
y  y 0  v 0 t  21 gt2


0  48m  0  21 9.8 m
t2
s
2
This time it is much easier to solve!
y=0
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
We get 2 answers again:
v = 0 here
t = 3.1s and t = -3.1s
Again we choose the positive value, but does it
actually answer our question?
v0 = 30 m/s
y=2.1m
y=0
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
v = 0 here
ymax=?
We get 2 answers again:
t = 3.1s and t = -3.1s
Again we choose the positive value, but does it
actually answer our question?
v0 = 30 m/s
y=2.1m
y=0
NO! This is only the time it takes to fall from the
maximum height. We need to add it to the time it
took to get to that max height. We found that
earlier – it was 3.1s
So our answer is the total time of 6.2s, as before.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
v = 0 here
ymax=?
Now we can do it one more time, but using
Equation (2) instead.
We already found the final speed to be v = 48m/s
Here is the calculation:
v  v 0  gt


 30.7 ms  30 ms  9.8 m
t
s
2
t  6.2s
v0 = 30 m/s
y=2.1m
y=0
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Sample Problem #2
A baseball is hurled straight up with an initial speed of 30 m/s. When it leaves the thrower’s hand the ball is 2.1 meters high.
Ignoring air resistance and assuming that the ball is allowed to fall to the ground, answer the following questions:
a)
What is the maximum height reached by the ball?
b)
When does it achieve that maximum height?
c)
What is the maximum speed achieved by the ball?
d)
When does the ball hit the ground?
v = 0 here
ymax=?
Now we can do it one more time, but using
Equation (2) instead.
We already found the final speed to be v = 48m/s
Here is the calculation:
v  v 0  gt


 30.7 ms  30 ms  9.8 m
t
s
2
t  6.2s
v0 = 30 m/s
y=2.1m
IMPORTANT NOTE: make sure you put in the
correct final velocity here. It is negative
because the ball is moving downward!
y=0
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Assistance Services at UCSB
Sample Problem #3
A model rocket is fired vertically from a launch pad on the ground. The engine provides an
upward acceleration of 5 m/s2 for a time of 10 seconds. After that time the rocket is in free fall.
Assume that the parachute fails to deploy, and the rocket eventually plummets to the ground.
Sketch graphs of the rocket’s vertical height vs. time and its velocity vs. time.
Find the maximum height of the rocket.
When does it land, and how fast is it moving just before it hits the ground?
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For Campus Learning
Assistance Services at UCSB
Sample Problem #3
A model rocket is fired vertically from a launch pad on the ground. The engine provides an
upward acceleration of 5 m/s2 for a time of 10 seconds. After that time the rocket is in free fall.
Assume that the parachute fails to deploy, and the rocket eventually plummets to the ground.
Sketch graphs of the rocket’s vertical height vs. time and its velocity vs. time.
Find the maximum height of the rocket.
When does it land, and how fast is it moving just before it hits the ground?
position
velocity
400
60
350
40
300
20
250
-20
150
-40
50
-60
0
-80
-50
0
0.8
1.6
2.4
3.2
4
4.8
5.6
6.4
7.2
8
8.8
9.6
10.4
11.2
12
12.8
13.6
14.4
15.2
16
16.8
17.6
18.4
19.2
20
20.8
21.6
22.4
23.2
100
position
0
0.8
1.6
2.4
3.2
4
4.8
5.6
6.4
7.2
8
8.8
9.6
10.4
11.2
12
12.8
13.6
14.4
15.2
16
16.8
17.6
18.4
19.2
20
20.8
21.6
22.4
23.2
0
200
-100
velocity
Answers:
Max Height = 378 m
Time to impact = 23.9 s
Impact speed = 86 m/s
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Assistance Services at UCSB