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?

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aristocles aristocles · Answer 2 · 2019-09-25T03:31:25+02:00

Answer:

Part a)

[tex]t = \frac{h}{v_o}[/tex]

Part b)

[tex]H = \frac{2v_o^2}{g}[/tex]

Part c)

[tex]H = \frac{v_o^2}{g}[/tex]

Explanation:

Part a)

Distance between two balls when it is dropped is given as

[tex]d = h[/tex]

relative velocity of two balls initially at time of drop

[tex]v_{rel} = v_o - 0[/tex]

relative acceleration of two balls is given as

[tex]a_{rel} = g - g = 0[/tex]

now we have

[tex]h = v_o t[/tex]

[tex]t = \frac{h}{v_o}[/tex]

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

[tex]t = \frac{2v_o}{g}[/tex]

Now if other ball just reach the ground

[tex]H = \frac{1}{2}gt^2[/tex]

so we have

[tex]H = \frac{1}{2}g(\frac{2v_o}{g})^2[/tex]

[tex]H = \frac{2v_o^2}{g}[/tex]

Part c)

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

then at the highest position

[tex]t = \frac{v_o}{g}[/tex]

this must be time of collision so we have

[tex]\frac{v_o}{g} = \frac{H}{v_o}[/tex]

[tex]H = \frac{v_o^2}{g}[/tex]