Question

A stone dropped from a bridge strikes the water 2.2 s later. How high is the bridge above thewater

Answer 1

Distance = ut+1/2gt^2
u is initial velocity g acceleration due to gravity t is time in seconds

u=0, t=2.2 g=9.8
distance = 0.5(9.8)(2.2)^2=23.716

Answer 2

Let the only force acting on the ball is the gravity (free-failing) and the ball was initially stationary
vo = 0

s = vot + 1/2 gt^2

s = 1/2 x 10 x (2.2) ^2 = 24.2 m