Question

Chan released a stone from a cliff of 10m height.

a)Determine the time taken for the stone to reach the bottom of the cliff.
b)Determine the velocity of the stone just before it touches the ground ignore air resistance.(g＝9.81ms-2)

Thanks​

Answer 1

a) time taken = 1.427 s

b) final speed before hitting the ground = 14 m/s

Step-by-step explanation:

height of cliff = 10 m

acceleration due to gravity = 9.81
`m/s^(2)`

time taken for stone to reach bottom of cliff = ?

velocity of stone just before hitting the ground = ?

To find the time taken, we will use Newton's equation of motion

using

`v^(2) = u^(2) + 2as`

where

v = final speed of stone before hitting the ground

u = initial speed of the stone (u = 0 since the stone started falling from rest)

a = acceleration due to gravity = 9.81
`m/s^(2)`

s = height of the cliff = 10 m

Solving, we have

`v^(2) = 0^(2) + 2(9.81 * 10)`

`v^(2) = 2*98.1`

`v^(2) = 196.2`

`v=√(196.2) = 14 m/s`

From the first equation of motion,

`v = u + at`

where t = time taken for stone to reach bottom of cliff

imputing values, we have

`14 = 0 + (9.81 * t)`

`t = (14)/(9.81) = 1.427 s`