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1st Equation of Motion

A stone proje...

Question

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

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

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

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

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g t_{1} t_{2}

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Solution

The correct option is C $\frac{1}{2} g t_{1} t_{2}$
$y = u t_{1} - \frac{1}{2} g t_{1}^{2}$
$\frac{t_{1} + t_{2}}{⇓} = \frac{2 u}{g}$
(total time of journey)
So, $u = g \frac{(t_{1} + t_{2})}{2}$
$\Rightarrow y = g \frac{(t_{1} + t_{2})}{2} t_{1} - \frac{1}{2} g t_{1}^{2} = \frac{1}{2} g t_{1} t_{2}$

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