Question

A rock is tossed straight up with a velocity of When it returns, it falls into a hole deep. What is the velocity as it hits the bottom of the hole?

Answer 1

Alright, let's say the rock is tossed up with an initial velocity of x. There is a downwards acceleration of -9.8ms^-2. I assume no air resistance. Let's say it falls into a hole L deep. Let's also say the rock is thrown from ground level, since you're not giving me any information.

x-x_0=v_0t+0.5gt^2
L=xt+0.5(-9.8)t^2

Solve for t. Once you find t, which should be a simple application of the quadratic formula, plug that t in for v=v_0+at, or v=x+(-9.8)t.

There. I can't give you any numbers, since you didn't give me any.