Question

a body dropped from the top of a cliff reaches the ground in 5 seconds. Find the height of the cliff​

Answer 1

Given :

time (t)= 5 secs.

acceleration due to gravity (g)= 9.77m/s^2

therefore , a= g

As it is dropped from height it's ,

initial velocity = 0 m/s

To find :

height of cliff =?

Formula :

s = ut +1 gt^2

2

Solution :

s = 0 ×5 + 1 ×9.77×5^2

2

s= 0 + 1 × 9.77 × 25

2

s = 9.77×25

2

s = 244.25

2

s= 122.125

s = 122.13 m

therefore , the height of cliff is 122.13 m

PLEASE MARK ME AS BRAINLIEST!!!

Answer 2

Answer:

125 meters (assuming g = 10m/s-²)

Explanation:

as per second equation of motion s = ut + 1/2at²

a = g = 10m/s^-2

t = 5 seconds

initial velocity = 0m/s

s = ut + 1/2at²

s = 0 * 5 + 1/2 * 10 * 5 *5

s = 0 + 5 * 5 * 5

s = 125

Note :- we have assumed g as 10 not 9.7 your answer could vary using 9.7