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A stone projected vertically up from the ground reaches a height y in its path at $t_{1}$ second and after further $t_{2}$ seconds reaches the ground. The height y is equal to

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$\frac{1}{2} g (t_{1} + t_{2)}$

$\frac{1}{2} g {(t_{1} + t_{2})}^{2}$

$\frac{1}{2} g t_{1} t_{2}$

$g t_{1} t_{2}$

answer is C.

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Detailed Solution

y=ut₁− $\frac{1}{2}$ g $t_{1}^{2}$

total time of journey (t₁ + t₂ ) = $\frac{2 u}{g}$

So, u = g ( $\frac{t_{1} + t_{2}}{2}$ )

y = g $\frac{t_{1} + t_{2}}{2} t_{1} - \frac{1}{2} g t_{1}^{2} = \frac{1}{2} g t_{1} t_{2}$

Therefore the value of y = $\frac{1}{2} g t_{1} t_{2}$

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