Question

A rock is tossed straight up with a speed of 23 m/s. When it returns, it falls into a hole 10 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

Height reached above the ground by the rock, h = u^2/(2*g)

= 23^2/(2*9.8)

= 27 m

distance travelled in downward direction,
H = h + 10

= 27 + 10

= 37 m

so, speed of rock at the bootom of the holw, v = sqrt(2*g*H)

= sqrt(2*9.8*37)

= 26.93 m/s <<<<<------Answer

Total time taken, T = t_up + tdown

= u/g + sqrt(2*H/g)

= 23/9.8 + sqrt(2*37/9.8)

= 5.1 s <<<<<------Answer