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# Sarah throws a ball to her friend Julie. What is the speed of the ball...?

Sarah throws a ball to her friend Julie. However, Sarah throws it too hard so it is over Julie's head when it reaches Julie's horizontal position. Assume the ball leaves Sarah's hand a distance 1.5 meters above the ground, reaches a maximum height of 8 m above the ground, and takes 1.44 s to get directly over Julie's head.

1) What is the speed of the ball when it leaves Sarah's hand?

2) How high above the ground will the ball be when it gets to Julie?

• When the ball reaches its maximum height, its vertical velocity is 0 m/s. The vertical acceleration is -9.8 m/s^2. Let’s use the following equation to determine the ball’s initial vertical velocity.

vf^2 = vi^2 + 2 * a * d, vf = 0, a = -9.8
0 = vi^2 + 2 * -9.8 * 8
vi = √156.8

This is approximately 12.5 m/s. Let’s use the following equation to determine the time for this to happen.

vf = vi – g * t
0 = √156.8 – 9.8 * t
t = √156.8 ÷ 9.8

This time is approximately 1.28 seconds. This is less than 1.44 seconds.
t = 1.44 – √156.8 ÷ 9.8

This time is approximately 0.162 second. Let’s use the following equation to determine the height of the ball at this time.

hf = hi + vi * t + ½ * g * t^2, hi = 8 meters, vi = 0 m/s, g = -9.8 m/s^2
hf = 8 + ½ * -9.8 * (1.44 – √156.8 ÷ 9.8)^2
This approximately 7.87 meters.
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• Edit
The ball reaches a height h = 6.5 m above from the throwing position, so its initial VERTICAL speed is sqrt(2gh) = 11.3 m/s using conservation of energy.

Horizontal velocity of the ball is of course unchanged throughout the flight, 23.057 m / 1.44 s = 16 m/s
So the total initial speed is sqrt(11.3^2 + 16^2) m/s = 19.59 m/s

b)
The height above ground of the ball as a function of time is
h = 11.3t - (9.8/2)t^2 + 1.5
Using the displacement equation for uniformly accelerated motion.
Where h is in m and t in s.
Plug in t=1.44 s.
h = 7.59 m
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