Question

A stone is dropped from the top of a 45 m high building .how fast will, it be moving when it reaches the ground ?And what will its velocity be?

Answer 1

s = 1/2 a t^2

t = under root of 45 x 2 / 10
t = root of 90 / 10
t = root of 9
t= 3 sec .

velocity = s / t
velocity = 45 / 3
velocity here = 15 m/s

the time it take is approx 3 sec and velocity will be 15 m/s

Answer 2

distance(s) = 45m
inital velocity (u)= 0
acceleration (a) = 9.8m/s^2
final velocity = v

using third eq. of motion

2as = v^2 -u^2
2(9.8)(45)= v^2 -(0)^2
882 = v^2
v = root 882
= 21 root 2