A ball projected from ground vertically upward is at same height at time $t_{1}$ and $t_{2}$ . The speed of projection of ball is: [Neglect the effect of air resistance].

g [t_{2} - t_{1}]

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\frac{g [t_{1} + t_{2}]}{2}

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\frac{g [t_{1} - t_{2}]}{2}

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g [t_{1} + t_{2}]

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Solution

The correct option is B $\frac{g [t_{1} + t_{2}]}{2}$
Let us say, the ball take $Δ t$
time to reach at to most point $3$ from point $1$
The ball will take time $Δ t$ again to reach from portion $3$ to portion $2$
$\Rightarrow t_{1} Δ t + Δ t = t_{2}$
$\Rightarrow t_{1} + 2 Δ t = t_{2}$
$\Rightarrow Δ t = \frac{t_{2} - t_{1}}{2}$
Let the projected speed of ball be it at position $3$ the speed of ball $= 0$
Applying $I$ equation of motion
$o = v - g (t_{1} + Δ t)$
$\Rightarrow v = g (t_{1} + Δ t)$
$v = g (t_{1} + \frac{t_{2} - t_{1}}{2})$
$\Rightarrow v = g \frac{(t_{1} + t_{2})}{2}$

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