Question

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

going when it hit the bottom of the cliff?

Answer 1

Step-by-step 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* 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.

Answer 2

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

Step-by-step 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