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  • 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.

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.

Answers (1)

For the first ball-

U = 0, h = 45m

a = g, t=t \; and\; V = v_{1} =?,

we know that,

v = u + at

v1 = 0 + gt = gt\; downward

(for second ball)

V = v_{2}

u = 40 m/s, a = -g,

t =t

now, V = u + at

v_{2} = (40 - gt) \; upward

Now relative velocity of the 1st ball w.r.t. 2nd

v_{12} = v_{1} - v_{2} = -gt -(40 - gt) = - gt - 40 + gt = - 40\; m/s

The speed of one ball increases and the speed of the other decreases with the same rate due to acceleration.

Hence, relative speed = 40 m/s.

Posted by

infoexpert22

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