Question

A man stands at the edge of a cliff and throws a rock downward with A speed of 12.0 m s . Sometimes later it strikes the ground 110 m below the place where it was thrown. A) how long does it take to reach the ground? B) what is the speed of the rock at impact?

Answer 1

B) 48.0 m/s

We can actually start to solve the problem from B for simplicity.

The motion of the rock is a uniformly accelerated motion (free fall), so we can find the final speed using the following suvat equation

`v^2 -u^2 = 2as`

where

`v` is the final velocity

`u = 12.0` is the initial velocity (positive since we take downward as positive direction)

`a=g=9.8 m/s^2` is the acceleration of gravity

s = 110 m is the vertical displacement

Solving for v, we find the final velocity (and so, the speed of the rock at impact):

`v = √(u^2+2as)=√(12^2+2(9.8)(110))=48.0 m/s`

A) 3.67 s

Now we can find the time of flight of the rock by using the following suvat equation

`v=u+at`

where

`v = 48.0` is the final velocity at the moment of impact

`u = 12.0` is the initial velocity

`a=g=9.8 m/s^2` is the acceleration of gravity

t is the time it takes for the rock to reach the ground

And solving for t, we find

`t=(v-u)/(a)=(48.0-12.0)/(9.8)=3.67 s`