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):

$v = \sqrt{u^2+2as}=\sqrt{12^2+2(9.8)(110)}=48.0 m/s$

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

$t=\frac{v-u}{a}=\frac{48.0-12.0}{9.8}=3.67 s$