Algebra Word Problem
Projectile Motion Problem
You know that a dropped object falls a distance
of 16t2 feet in t seconds. When an object is not
simply released but is thrown or launched, it is
called a projectile. What happens to a
projectile that is launched with some initial
vertical velocity, Vo (measured in feet per
second), at some initial height ho (measured in
feet)?
• Without gravity to pull the
projectile its height h Would
increase according to the
equation h = vo t + ho
• With gravity, the projectile
falls 16t2 feet in t seconds.
• So the projectile’s height at
any time t is given by
H = -16t2 + v0t + h0
In order to solve quadratic equations
involving maximums and minimums for
projectile motion, it is necessary to
• know how to solve quadratic equations:
• know vertex formula for a parabola
• write and solve an equation for the
problem
Let’s solve the example of a quadratic equation
involving maximums and minimums for projectile
motion
• An object is launched at 19.6 meters
per second (m/s) from a 58.8-meter tall
platform. The equation for the object's
height s at time t seconds after launch
is s(t) = –4.9t2 + 19.6t + 58.8, where s is
in meters. When does the object strike
the ground?
• What is the height (above ground level)
when the object smacks into the ground?
• Well, zero, obviously.
• So I'm looking for the time when the height
is s = 0.
• I'll set s equal to zero, and solve:
• 0 = –4.9t2 + 19.6t + 58.8
• 0 = –4.9t2 + 19.6t + 58.8
0 = t2 – 4t – 12
0 = (t – 6)(t + 2)
Then t = 6 or t = –2.
• The second solution is from two seconds
before launch, which doesn't make sense
in this context.
• So "t = –2" is an extraneous solution, and
I'll ignore it.
Answer:
The object strikes the ground six
seconds after launch.
Physics Exploration
You’re playing Angry Birds on your sweet new
iPhone and get curious about the maximum
height, distance, and time the bird is in the air
before it hits the ground again. This formula is
from Science class :
s(t )  gt  v0t  h0
2
s(t )  gt  v0t  h0
2
Key Information Box:
t = Time (in seconds)  Our unknown
variable in most cases
g = Gravity  4.9 if unit is meters, 16 if unit
is feet
V0 = Initial velocity  either m/s or ft/s
h0 = Initial Height of objectfeet or meters
s(t )  gt  v0t  h0
2
If you launch a bird at an initial velocity of 10
m/s from the ground, how long will it take until
the bird hits the ground again?
What do you know?
•v0= 10 m/s
• g = 4.9 (because we’re in meters)
• h0= 0 because FROM GROUND
s(t )  gt  v0t  h0
2
If you launch a bird at an initial velocity of 10
m/s from the ground, how long will it take until
the bird hits the ground again?
What do you WANT to know?
• time until s(t) = 0 again.
0  4.9t  10t  0
2
0  4.9t  10t  0
2
• A = -4.9
• B = 10
• C=0
• Answer: Hits ground again 2.04 sec. after
launch.
• Key Question: Why not the negative?
Another Problem:
• An object in launched directly upward
at 64 feet per second (ft/s) from a
platform 80 feet high. What will be the
object's maximum height? When will it
attain this height?
• Hmm... They didn't give me the equation
this time. But that's okay, because I can
create the equation from the information
that they did give me.
• The initial height is 80 feet above ground
and the initial speed is 64 ft/s. Since my
units are "feet", then the number for
gravity will be 16, and my equation is:
• s(t) = –16t2 + 64t + 80
• They want me to find the maximum height.
For a negative quadratic like this, the
maximum will be at the vertex of the
upside-down parabola. So they really want
me to find the vertex.
t = –b/2a = –(64)/2(–16) = –64/–32 = 2
s = s(2)
= –16(2)2 + 64(2) + 80
= –16(4) + 128 + 80 = 208 – 64 = 144
• And in this case, the vertex is at (2, 144).
• But what does this vertex tell me?
According to my equation, I'm plugging in
time values and extracting height values,
so the input "2" must be the time and the
output "144" must be the height.
• It takes two seconds to reach the
maximum height of 144 feet
New Question
If you launch a bird at an initial velocity of 10
m/s from the ground, what is the maximum
height the bird will reach? When will the
bird reach this height?
What do we want to know?
• VERTEX!
y  4.9t  10t  0
2
• Graph it and find the vertex.
• Max height is y-coordinate, time is xcoordinate.
• In a sentence: “Bird reaches maximum
height of 5.16 meters after 1.06 seconds.
Now… Practice!
An object in launched directly upward at 64 feet per
second (ft/s) from a platform 80 feet high.
• Question 1: What is the initial velocity (v0)?
• Question 2: What is the initial height (h0)?
• Question 3: What is the value for g?
• Question 4: What’s our equation?
• Question 5: What are the values for a, b, and c?
• Question 6: How long will it take before the object hits
the ground?
• Question 7: What is the maximum height, and after how
many seconds will the object reach that height?