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

Part a)

`t = (h)/(v_o)`

Part b)

`H = (2v_o^2)/(g)`

Part c)

`H = (v_o^2)/(g)`

Step-by-step 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 = (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 = (2v_o)/(g)`

Now if other ball just reach the ground

`H = (1)/(2)gt^2`

so we have

`H = (1)/(2)g((2v_o)/(g))^2`

`H = (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 = (v_o)/(g)`

this must be time of collision so we have

`(v_o)/(g) = (H)/(v_o)`

`H = (v_o^2)/(g)`

Answer 2

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.