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

We can start with the displacement equation for constant acceleration (in the terms of the variables given):
h = vi*dt + 0.5*g*dt^2

The initial velocity of the dropped 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)

plug equation (3) into equation (2) and solve for vi.

h = vi*(sqrt(2*h/g) - t) + 0.5*g*(sqrt(2*h/g) - t)^2

vi = ( h - 0.5*g*(sqrt(2*h/g) - t)^2 ) / (sqrt(2*h/g) - t)

that is simple relation for intial velocity of second stone in terms of h, g , and t .