A rubber ball is shot straight up from the ground with speed v0. Simultaneously,

Question

A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. (a) At what height above the ground do the balls collide? (Use the following as necessary: g, h, and v0.) ycoll = (b) What is the maximum value of h for which a collision occurs before the first ball falls back to the ground? (Use the following as necessary: g, and v0.) h = (c) For what value of h does the collision occur at the instant when the first ball is at its highest point? (Use the following as necessary: g, and v0.) h =

Explanation / Answer

Here,

Let ground level be 0, and let the "up" direction be positive. If U(t) is the displacement at time t of the ball that was thrown up, then

U(t) = v0*t - 0.5 * g*t^2

until the ball returns to ground. If D(t) is the displacement at time t of the ball that was dropped, then

D(t) = h - 0.5*g*t^2

b)

The first ball returns to the ground at time t=2*v0/g. For the maximum height h so that the two balls collide before the first hits the ground. There is technically no such maximum height.But for h so that they collide exactly when the first (and second) hits the ground. Just set the t we found in part (a) equal to 2v0/g.

c)

The ball thrown up reaches is highest point at t=v0/g.

S = 0.5*g* t^2 = 0.5* g*(v0/g) ^2 = v0^2/2g

h = 2 *S = v0^2/g

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