Question

A body dropped from the top of a tower reaches the ground in 4s. I light of the
39.2 m
58.8 m
44.1 m
78.4 m
​

Answer 1

Answer:

78.4m

Explanation:

apply second equation of motion

$s = ut + \frac{1}{2} g{t}^{2}$

when a body will be dropped from any height then it's initial velocity will be always zero

so u=0

then

$s = 0 \times t + \frac{1}{2} \times 9.8 \times {4}^{2} = 78 .4m$