Master

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 thehorizontal.
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 initialvelocity.
v x = vi cos q = (4.36m / s)(cos 81) = 0.6821m / s
viy = vi sin q = (4.36m / s)(sin 81) = 4.3063m / s
1
y = viy t + 2 at 2

†1
0m = (4.3063m / s) t + 2 (-9.81m / s2 ) t 2

0m = (4.3063m / s - 4.905m / s2 t ) t
t = 0s
0m = 4.3063m / s - 4.905m / s2 t4.3060m / s
t=
4.905m / s2
t = 0.8779s

It took the ball 0.88 seconds to land.
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b) How high did the ball bounce?

Solution:Because the take-off and landing heights are the same, the parabola is
completely symmetrical and the maximum height will occur at halfthe flight
time. Therefore it will occur when time is 0.4389 seconds.
1
y = viy t + 2 at 2
1
y = (4.3060m / s)(0.4389s) + 2 (-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.

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c) What was theball's range?
Solution:
x = vxt
x = (0.6821m / s)(0.8779s)
x = 0.5988m

The ball will travel 0.60 meters down range.
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