A ball is dropped from the top of a building at $t = 0$ . At a later time $t = t_{0}$ , a second ball is thrown downward with an initial speed $v_{0}$ . Obtain an expression for the time t at which the two balls meet.

$[\frac{v_{0} - \frac{g t_{0}}{2}}{v_{0} - g t_{0}}] t_{0}$

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$[\frac{v_{0} + \frac{g t_{0}}{2}}{v_{0} - g t_{0}}] t_{0}$

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$[\frac{v_{0} - \frac{g t_{0}}{2}}{v_{0} + g t_{0}}] t_{0}$

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$[\frac{v_{0} + \frac{g t_{0}}{2}}{v_{0} + g t_{0}}] t_{0}$

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Solution

The correct option is A
$[\frac{v_{0} - \frac{g t_{0}}{2}}{v_{0} - g t_{0}}] t_{0}$

When the balls collide at time $t = t$ , the distance covered by them are same. But the caution is that the first ball has moved for $t$ time, while the second has moved for $(t - t_{0})$ time.
$∴ \frac{1}{2} g t^{2} = v_{0} (t - t_{0}) + \frac{1}{2} g (t - t_{0})^{2}$
After simplifying, we get, $t = [\frac{v_{0} - \frac{g t_{0}}{2}}{v_{0} - g t_{0}}] t_{0}$

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