Question

A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone? Give your answer in terms of the given variables and g.

Answer 1

The equation for displacement of a particle moving with constant acceleration is:
   

       h = (vi * dt) + (0.5 * g * dt^2)

   The initial velocity of the dropped stone(first stone) is 0m/s; hence for the dropped stone:
    h = 0.5 * g * dt^2 --->(1)

For the second stone the duration of time it takes to reach the water would be dt - t
Hence for the second stone:

h = [vi * (dt - t)] + [0.5 * g * (dt - t)^2 ] ---> (2)

We want vi in terms of h, g, and t only, so we need to get rid of the dt term. To do this, solve equation (1) for dt to get:
  

   dt = sqrt(2*h/g) ---> (3)

from equation (2) we get

  

vi = [1/(dt - t)] [h - (0.5) g (dt - t)^2] -> this equation is interms of time taken by first stone, dt.

plug equation (3) into above equation to further simplify this to get

vi = {h - (0.5) g [(2h/g)^(1/2) - t]^2} / {(2h/g)^(1/2) - t}

(You can rearrange the above equation in required format)