Question #aeb96

1 Answer
Jan 29, 2018

therefore The height of the cliff is 62.145 m

Explanation:

Let initial velocity be given by 'u' = 6.0ms^-1,
and
Final velocity be given by 'v' = 0 ms^-1.
't' = 3.0s is the time taken by the stone to reach the ground, and
'a' = 9.81ms^-2 is the acceleration due to gravity.
Let 's' be the distance covered.
According to the position- time relationship or the second equation of motion:

s=ut +1/2at^2

s = 6.0 m/cancels xx 3.0cancels + 1/2 xx 9.81m/cancels^2 xx (3.0)^2 cancels^2

s = 18 m + (9.81 xx 9)/2 m

s= 18m + 44.145m

s= 62.145 m is the distance covered by the stone from top of the cliff to the bottom.

therefore The height of the cliff is 62.145 m