Aball dropped from a bridge takes 3.0 seconds to reach the water below. how far is the bridge from the water?

Since it is dropped, initial velocity u = 0

Using s = ut + (1/2)gt^2,        putting u = 0,  g = 10 m/s^2

s =(1/2)gt^2,                  t = 3s

s = 0.5 * 10 * 3 *3

s = 45 m.

The bridge is 45m above the water.

The answer is as follows in the photo