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

20 m s^{- 1}

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40 m s^{- 1}

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30 m s^{- 1}

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0 m s^{- 1}

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Solution

The correct option is B $40 m s^{- 1}$
According to the problem, for the ball dropped from the building, $u_{1} = 0$ and $u_{2} = 40 m / s$

Velocity of the ball after time $t$ ,

$v_{1} = u_{1} - g t$

$v_{1} = - g t$

And for another ball which is thrown upward,

$u_{2} = 40 m / s$

Velocity of the ball after time $t$ ,

$v_{2} = u_{2} - g t = (40 - g t)$

Therefore,

Relative velocity of one ball with respect to another ball is,

$v_{12} = v_{1} - v_{2} = - g t - [- 40 - g t]$

$v_{12} = v_{1} - v_{2} = - g t + 40 + g t = 40 m / s$

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

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