Question

A rubber ball is shot straight up from the ground with speed v_0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. 1.) At what height above the ground do the balls collide in terms of v_0, g, and h? 2.) What is the maximum value of h for which a collision occurs before the first ball falls back to the ground in terms of v_0? 3.) For what value of h does the collision occur at the instant when the first ball is at its highest point in terms of v_0?

Answer 1

The following are the answers to this specific problem:              

1. h−gh2 / 2v20

2. h(max) = Vo^2/g

3. h(max) = Vo^2/g

I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

Answer 2

Answer:

Part a)

$t = \frac{h}{v_o}$

Part b)

$H = \frac{2v_o^2}{g}$

Part c)

$H = \frac{v_o^2}{g}$

Explanation:

Part a)

Distance between two balls when it is dropped is given as

$d = h$

relative velocity of two balls initially at time of drop

$v_{rel} = v_o - 0$

relative acceleration of two balls is given as

$a_{rel} = g - g = 0$

now we have

$h = v_o t$

$t = \frac{h}{v_o}$

Part b)

now the maximum height that we can use here is H

then we can say time to reach the ball back on the ground which is projected from ground is

$t = \frac{2v_o}{g}$

Now if other ball just reach the ground

$H = \frac{1}{2}gt^2$

so we have

$H = \frac{1}{2}g(\frac{2v_o}{g})^2$

$H = \frac{2v_o^2}{g}$

Part c)

If collision occurs at the top position of ball projected from the ground

then at the highest position

$t = \frac{v_o}{g}$

this must be time of collision so we have

$\frac{v_o}{g} = \frac{H}{v_o}$

$H = \frac{v_o^2}{g}$