Question:
A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed 40 m/s. Calculate the relative speed of the balls as a function of time.
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