Question

A body dropped from the top of a building takes 4seconds to reach the ground.The height of the building is:

Answer 1

u = 0 m/s as when a body is dropped its initial velocity is zero
g (acceleration due to gravity) = 10m/s²
t = 4 sec
s = height of building
using 2nd law ,
s = ut+1/2gt²
s = 0(4)+1/2(10)(4)²
s = 0 + 160/2
s = 80 m
hope this helps

Answer 2

u = 0

Height of the building
H = 0.5gT^2
= 0.5 * 9.8 * 4^2
= 4.9 * 16
= 78.4 m

Height of the building is 78.4 m