Solution to Example from Module 5:
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?
Known:
Up will be considered positive and down will be considered negative
a = g = -9.81 m/s2
vi = 4.36 m/s @ 81°
y=0m
t=?
Solution:
First, determine the horizontal and vertical components of the initial velocity.
v x = vi cos q = (4.36m / s)(cos 81) = 0.6821m / s
viy = vi sin q = (4.36m / s)(sin81) = 4.3063m / s
y = viy t + 21 at 2
†
0m = (4.3063m / s)t + 21 (-9.81m / s2 )t 2
0m = (4.3063m / s - 4.905m / s2 t)t
t = 0s
0m = 4.3063m / s - 4.905m / s2 t
4.3060m / s
t=
4.905m / s2
t = 0.8779s
It took the ball 0.88 seconds to land.
† high did the ball bounce?
b) How
Solution:
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 + 21 at 2
y = (4.3060m / s)(0.4389s) + 21 (-9.81m / s2 )(0.4389s)2
y = 1.8899m - 0.9449m
y = 0.9450m
The ball will reach a maximum height of 0.95 meters.
†
c) What was the ball's range?
Solution:
x = vxt
x = (0.6821m / s)(0.8779s)
x = 0.5988m
The ball will travel 0.60 meters down range.
†