Question

2. If a rock fell down a cliff and hit the bottom of the ravine at 4 seconds, how fast was the rock

Answer 1

Answer: -39.2 m/s or 39.2 m/s directed downwards

Explanation:

This situation is a good example of Free Fall, where the main condition is that the initial velocity must be zero $V_{o}=0$, and the acceleration is constant (acceleration due gravity).

So, in order to calculate the final velocity $V$ of the rock just at the moment it hitsthe bottom of the cliff, we will use the following equation:

$V={V_{o}}^{2}+gt$

Where:

$g=-9.8 m/s^{2}$ is the acceleration due gravity (directed downwards)

$t=4 s$ is the time it takes to the rock to fall down the cliff

$V=(-9.8 m/s^{2})(4 s)$

$V=-39.2 m/s$ This is the rock's final velocity and its negative sign indicates it is directed downwards

Answer 2

Explanation:

It is given that,

Initial speed of the rock, u = 0

It hits the bottom of the ravine at 4 seconds. Let v is the speed of the rock when it hits the bottom of the cliff. It will move under the action of gravity. Using equation of kinematics as :

$v=u+at$

a = g

$v=u+gt$

$v=gt$

$v=9.8\ m/s^2\times 4\ s$

v = 39.2 m/s

So, the speed of the rock when it hit the bottom of the cliff is 39.2 m/s. Hence, this is the required solution.