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A ball A is dropped from a building of ...

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.

Text Solution

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Speed of falling ball (A) after time ` t, v_(1) = gt (downwards )
Speed of projected upward ball (B) after time ` t, v_1 = (40 -gt) (upwards ) =- (40- gt) (downwards)
Relative velocity of ball (A) w.r.t. another velocity of ball ` B = v_1 -v_2 = gt -[-(40-gt0] =40 m//s`
At time ` t=0`, the relative velocity of ball ` A w.r.t ball (B), u_(AB) =0 - (-40) = ms^-1). Since ` v_(AB)=u_(AB)`, therefore, there is no acceleration of ball ` A w.r.t. ball B`. Hence the relative speed of ball (A) at any instant of time remains constant .
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