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

Options

A. 0
B. `10 m s^(-1)`
C. `25 m s^(-1)`
D. `50 m s^(-1)`

Answer

D

Explanation

Here, `u_A = 0, u_B = +50 m s^(-1)`
`a_A = -g, a_B = -g`
`u_BA = u_B - u_A = 50 m s^(-1) - 0 m s^(-1) = 50 m s^(-1)`
`a_BA = a_B - a_A = -g - (-g) = 0`
since `v_BA = u_BA + a_Bat` (As `a_BA= 0`)
`therefore v_BA = u_BA`
As there is no acceleration of ball B w.r.t to ball A, therefore the relative speed of ball B w.r.t ball A at any instant of time remains constant `( = 50 m s^(-1))`