A ball falls from height $h$ . After $1$ second, another ball falls freely from a point $20 m$ below the point from where the first ball falls. Both of them reach the ground at the same time. What is the value of $h$ ?

11.2 m

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21.2 m

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31.2 m

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41.2 m

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Solution

The correct option is C $31.2 m$
Let the instant when first
ball is dropped be t 20s
and the time of fight
of first ball be t
Then,
the time of fight of second ball $= (b - 1) s$
To find the height,
Applying second $e q^{n}$ of motion so,
Fort ball $- 1, h = 0 (t) + \frac{g}{2} t^{2}$
$h = \frac{g}{2} t^{2} \Rightarrow t = \sqrt{\frac{2 h}{g}} s$
For ball $- 2,$
$(h - 20) = - 0 (t - 1) + \frac{g}{2} (t - 1)^{2}$
$h - 20 = \frac{g}{2} (t - 1)^{2}$
$h - 20 = \frac{g}{2} t^{2} + \frac{g}{2} - g t$
$\Rightarrow h - 20 = h + \frac{g}{2} - g \sqrt{\frac{2 h}{g}}$
$\Rightarrow h - 20 = h + \frac{g}{2} - \sqrt{2 h g}$
$\Rightarrow h = \frac{(\frac{g}{2} + 20)^{2}}{2 g}$
If $g = 10 m / s^{2}$
$h = \frac{(25)^{2}}{20} = 31.25 m$
Option - C correct.

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A ball falls from height h. After 1 second, another ball falls freely from a point 20m below the point from where the first ball falls. Both of them reach the ground at the same time. What is the value of h?

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