A body is projected vertically upwards. If $a$ and $b$ be the times at which it is at height $h$ above the point of projection while ascending and descending respectively, then $h$ is :
(Take $g = 10 m / s^{2}$ )

5 a b

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4 a b

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a b

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2 a b

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Solution

The correct option is A $5 a b$
If the body is projected vertically upwards with a velocity $u$ ,
$h = u t - \frac{1}{2} g t^{2}$ (using formula $S = u t + \frac{1}{2} a t^{2}$ )
or $t^{2} - (2 u / g) t + 2 h / g = 0$
This is a quadratic equation of $t$ and it has two roots $t_{1} = a$ and $t = b$ (say)
We know that, product of roots $=$ constant term/coefficient of $t^{2}$
So, $a b = (2 h / g) / 1$ or $h = g a b / 2$

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A body is projected vertically upwards. If a and b be the times at which it is at height h above the point of projection while ascending and descending respectively, then h is :(Take g=10 m/s2)

A body is projected vertically upwards. If $a$ and $b$ be the times at which it is at height $h$ above the point of projection while ascending and descending respectively, then $h$ is :
(Take $g = 10 m / s^{2}$ )