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Question

A boy standing on a bridge over a river 30.0 m below drops a stone straight down and finds that it takes 2.0 seconds to hit the water. Assume that the acceleration of the stone is 9.8 m/s² downwards. Find the initial velocity of the stone.

Answer 1

This is a one dimensional kinematics problem. We know the time, acceleration, and distance covered. We want to know the initial velocity. Here we would use the first kinematic equation.

y=y₀+v₀t+(1/2)at²

Since the rock hits the river we can consider the final position y=0, y₀=30 a=-9.8 m/s² t=2.0s

simplifying this we can rewrite this as

0=y₀+v₀*t-(1/2)gt² or v₀*t=(gt²)/2-y₀

Getting a common denominator we get

v₀t=(gt²-2y₀)/2 or v₀=(gt²-2y₀)/2t

Pluging in the numbers we get

v₀=(9.8*4-2*30)/(2*2)=-5.2m/s

or 5.2 m/s downward.

Answer 2

d = V₀*t + .5*a*t²

d = V₀*t - .5*9.8*t²

-30 = V₀*2 - 19.6

V₀ = -5.2 m/s or just say 5.2 m/s down