Physics 12: Projectiles launched at an angle
Please see the answers to the assigned problems 1-4 on textbook page 99 posted.
The key to projectiles launched at an angle is to split the problem into two: Horizontal and Vertical
components.
Remember that there will be no change in the horizontal component of its velocity (because no
acceleration)
When it reaches the highest point in its flight, it isn’t moving up, and it isn’t moving down, for an
instant of time… its vertical velocity is ZERO.
By the time it reaches the initial height it was launched, it will have just as much vertical velocity
as when it left. The vertical velocity will be equal and opposite to the initial velocity. Viy= -Vfy (at
that same height)
Again, remember through out the flight the horizontal velocity remains the same.
1st. Use trigonometry to split the initial velocity (the total initial velocity at which it is
launched at an angle) into the horizontal and vertical velocities.
Now your velocity and acceleration formulas will work…. ONLY LOOKING AT EITHER THE
HORIZONTAL OR THE VERTICAL RANGES. Don’t mix them up.
But there is no acceleration in the horizontal range.
So Vx=d/t
BUT In the vertical range there is acceleration
Vf2 = vi2 + 2ady
Vf=Vi +a(T)
Dy = (Vi)(T) + 0.5 (a) (T)2
Let’s look at an example:
(gravity)
You kick a soccer ball at an angle of 40° above the ground with a velocity of 20m/s.
How high will it go?
How much time does it spend in the air?
How far away from you will it hit the ground (horizontal distance)?
What is the ball’s velocity when it hits the ground?
Before we can calculate anything else, we first need to break the original velocity into
components.
vertical component
sinΘ = opp/hyp
opp = sinΘ (hyp)
sin 40° (20m/s) = vy = 13m/s
horizontal component
cosΘ = adj/hyp
adj = cos Θ (hyp)
cos 40° (20m/s)
= vy = 15m/s
a) TO FIND TIME we look at the VERTICAL!
At its maximum height, halfway through its flight, the object
won't be going up or down, so the velocity (we will call it
Vpeak) at that point is zero.
Vpeak2 = vif2 + 2ad d = (vf2 - vi2) / 2a
= (02 - 132) / 2(9.81)
d = 8.6 m
The ball will reach a maximum height of 8.6 m.
b) THINK VERTICAL!
Same ideas as above...
a = (vpeak - vi) / t t = (vpeak- vi) / a
= (0 - 13) / -9.81 t = 1.3 s
But this is only the time to the halfway point, so the final answer is 2.7s.
c) Now we can use the time to relate to the HORIZONTAL
There is no acceleration in the horiztonal range so
v = d/t
d = vt
= 15m/s (2.7s)
d = 41 m
41m is the horizontal distance (range) that it has traveled.
d) The ball’s velocity when it hits the ground is exactly the same as when it was originally
launched… 20 m/s at 40° up from the horizontal. The only difference is that now it's spiking into
the ground.
Example 2:
A bouncing ball leaves the ground with a velocity of 4.36 m/s at an angle of 81 degrees above the
horizontal. a) How long did it take the ball to land?
b) How high did the ball bounce?
c) What was the ball's range?
Solutions:
1st First, determine the horizontal and vertical components of the initial velocity.
vx = vi cosq = (4.36m/ s)(cos81) = 0.6821m/ s
viy = vi sinq = (4.36m/ s)(sin81) = 4.3063m/ s
a) Here is another way to find time….. the total displacement travelled in the vertical ( Dy) is zero.
y = viy t + 1/2 at2
0m= (4.3063)t + 1/2 (-9.81 )t2 factor out t
0m= (4.3063 - 4.90t)t divid both sides by t
0m= 4.3063 - 4.905t rearrange
t = 4.3060 - 4.905
t = 0.8779s
Because the take-off and landing heights are the same, the parabola is completely symmetrical
and the maximum height will occur at half the flight time. Therefore it will occur when time is
0.4389 seconds.
y = viy t + ½ at2
y = (4.3060)(0.4389) + 1/2 (-9.81)(0.4389)2
y = 1.8899m- 0.9449m
y = 0.9450m
The ball will reach a maximum height of 0.95 meters.
c) Solution:
Dx = (vx)t
Dx = (0.6821m/ s)(0.8779s)
Dx = 0.5988m
The ball will travel 0.60 meters down range.
EXAMPLE 3:
A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above
the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the
long-jumper.
First, use trigonometric function to determine the initial velocity components:
Horiz: vix =(12 m/s) • cos (28 deg) =10.6 m/s
Vert: viy =(12 m/s) • sin (28 deg) = 5.6 m/s
List the known information.
Horizontal Info:
x = ???
vix = 10.6 m/s
ax = 0 m/s/s
Vertical Info:
y = 0.0 m (total); ypeak = ???
viy = 5.6 m/s
vypeak = 0 m/s
vfy = -5.6 m/s
ay = -9.8 m/s/s
Use vfy = viy + ay • ttotal and vertical info to solve for time; t
-5.6 = 5.6 + -9.81 T
T= (-5.6 + -5.6) / (-9.81)
the time of flight is 1.1 seconds (rounded from 1.1497 s).
Now use x = vix • t
10.6 m/s for vix and 1.1 s for t, the x can be found to be 12.2 m (rounded from 12.1817 m)
Finally, use tup = 0.5748 s and the equation ypeak = viy • tup + 0.5 • ay • tup2 with vertical info to
find the y at the peak. Substituting and solving yields 1.6 m (rounded from 1.6193 m)