# you throw a ball straight upward into the air with a velocity of 20.0 m/s, and you catch the ball some time later.

a) how long is the ball in the air
b)how high does the ball go
c) what is the balls velocity when you catch it
I don't get how you are suppose to figure this out without the weight of the ball

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1. (B) How high does the ball go?
Since the ball is in uniformly accelerated motion, we can use the formula,
vf^2 - vo^2 = 2gh
where
vf = final velocity
vo = initial velocity
g = acceleration due to gravity
h = height

Since at maximum height, the ball stops moving (vf = 0). Thus,
0 - 20^2 = 2(-9.8)h
h = -400 / (-19.6)
h = ?

(C) What's the ball's velocity when you catch it?
Well, when you've caught it, it stops at your hands. So its velocity is zero.

(A) How long is the ball in the air?
After solving for the maximum height in (B), you can get the time using the formula,
h = vo*t - (1/2)gt^2

Just plug in the values in this formula and solve for t. After solving, multiply it by 2 to account for the ball falling back to the ground.

Hope this helps~ `u`

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2. 20

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