A ball is dropped from a building of height 45 m.

Solution:

V = v1 = ?

U = 0

h = 45 m

a = g

t = t

V = u + at

v1 = 0 + gt

v1 = gt

Therefore, when the ball is thrown upward, v1 = -gt

V = v2

u = 40 m/s

a = g

t = t

V = u + at

v2 = 40 – gt

The relative velocity of the ball in the downward direction is – 40 m/s

But when the speed increases due to acceleration, the relative speed remains 40 m/s