Mastering physics a rock tossed?

Question

A rock is tossed straight up with a speed of 23m/s. When it returns, it falls into a hole 14 m deep.

What is the rock's velocity as it hits the bottom of the hole?

How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

Answer 1

It's easiest to use conservation of energy for the first part.  Let vi=initial speed, v=final speed, d=depth of hole:

mvi^2/2=mv^2/2 - mgd

Thus

v=sqrt(vi^2 + 2 gd)=28.34 m/s.

Let t1 be the time to reach the highest point.  This time is the same as if you were to drop the rock from that place to hit the ground:

vi=g t1

Thus t1=v1/g=2.35 s.The time to reach the bottom of the hole (measured from the highest point) is

t2=v/g=28.34/9.8=2.89 s.

The total time is t1+t2=2.35+2.89=5.24  s.

Answer 2

Find the max height above ground level H = v^2/2g; where v = 23 mps up and g = 9.81 m/sec^2.

Velocity at the bottom of the hole = V^2 = 2g(H + d); where d = 14 m.

T = tup + tdown the total time in the air.  tup^2 = 2H/g.  tdown^2 = 2(H + d)/g.  Thus T = sqrt(2H/g) + sqrt(2(H + d)/g).

You can do the math.