Question

A ball A is dropped from a building of height 45 m, Simultaneously another identical ball B is thrown up with a speed 50 m s^-1. The relative speed of ball B w.r.t. ball A at any instant of time is (Take g = 10 m s^-2)

• A. 0 m s^-1
• B. 10 m s^-1
• C. 25 m s^-1
• D. 50 m s^-1

Answer 1

Answer: (D)

50 m s^-1

Answer 2

Here, $u_{A} = 0$

$\hat { \gamma } ^ { + v e }$

$u_{B} = + 50ms^{- 1}$

$a_{A} = - g,a_{B} = - g$

$u_{BA} = u_{B} - u_{A} = 50ms^{- 1} - 0ms^{- 1} = 50ms^{- 1}$

$a_{BA} = a_{B} - u_{A} = - g - ( - g) = 0\because v_{BA} = u_{BA} + a_{BA}t(Asa_{BA} = 0)$

$\therefore v_{BA} = u_{BA}$

As there is no accelerations 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 ($= 50ms^{- 1}$)