Physics 111
Projectile Motion 2.0
Practice Problem

A rock is thrown with an initial vertical velocity
component of 30 m/s and an initial horizontal
velocity component of 40 m/s.
◦ What will these velocity components be one
second after the rock reaches the top of its
path?
◦ Assuming the launch and landing heights are
the same, how long will the rock be in the air?
◦ Assuming the launch and landing heights are
the same, how far will the rock land from
where it was thrown?
Solution:

A. The horizontal component of the
velocity remains constant, 40 m/s.

The vertical component of the velocity
decreases by 10 m/s every second. So at
the peak, the vertical component of the
velocity is zero and one second later the
vertical component of the velocity is -10
m/s.
Solution:

b. Assuming the launch and landing heights
are the same, how long will the rock be in
the air?
Since the vertical component of the velocity
decreases by 10 m/s every second, it will
take 3 seconds for the vertical component
of the velocity to slow from 30 m/s to 0 m/s,
and another 3 seconds for the rock to
accelerate from 0 m/s to -30 m/s. The total
time in the air is 6 seconds.
Solution:

c. Assuming the launch and landing
heights are the same, how far will the
rock land from where it was thrown?
Since the rock is in the air for six seconds
and the rock moves horizontally 40
meters each second, the range for the
rock is 6 s x 40 m/s = 240 m.
Practice Problem 2

A pool ball leaves a 0.60-meter high table
with an initial horizontal velocity of 2.4
m/s. Predict the time required for the
pool ball to fall to the ground and the
horizontal distance between the table's
edge and the ball's landing location.
Solution:

Time:
◦ Use delta y = Vyi(t) + .5(a)(t^2)
◦ (0-.60) = 0(t) + .5(-9.8)(t^2)
◦ T = .17 seconds
Delta x
◦ Use equation delta x= .5( vf + vi)(t)
◦ = .5(0+24)(.17)
◦ = 2.1 meters
Problem 3
A long jumper leaves the ground with an
initial velocity of 12 m/s at an angle of 28degrees above the horizontal.
 Determine:

◦ time of flight
◦ the horizontal distance
◦ peak height of the long-jumper
Solution:

Time:
◦ Use vf= vi + at ( because all others have delta y which
we don’t have)
◦ 0 = 3.71 + (-9.8)(t)
◦ T= .37 seconds

Delta x
◦
◦
◦
◦

Use equation Delta x = .5 ( vf + vi)(t)
Because it doesn’t have acceleration and has time
(xf – 0)= .5 ( 0 + 11.41)(.37)
= 2.11 meters
Delta y / 2
◦
◦
◦
◦
Max height will be mid flight or .37/2 = .185 seconds
Use equation Delta y = Vi(t) + .5(-9.8)(t^2)
Delta y = 3.71 ( .185) + .5 (-9.8) (.815^2)
Delta y for half way or mazx height is .52 meters
Practice 4

Alice throws the ball to the +X
direction with an initial velocity 10m/s.
Time elapsed during the motion is 5s,
calculate the height that object is thrown
and Vy component of the velocity after it
hits the ground.
Answer:
Problem

John kicks the ball and ball does
projectile motion with an angle of 53º to
horizontal. Its initial velocity is 10 m/s, find
the maximum height it can reach,
horizontal displacement and total time
required for this motion.
Anwser: