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A ball is thrown vertically upwards with a velocity of $19 .6 {ms}^{- 1}$ from the top of a tower. The ball strikes the ground after 6s. The height from the ground up to which the ball can rise will be $(\frac{k}{5}) m .$ The value of k is _______(use $g = 9.8 m / s^{2}$ )

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answer is 392.

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Detailed Solution

$s = ut + \frac{1}{2} {at}^{2}$

Let say height of tower is h.

$\Rightarrow - h = (19.6 \times 6) - \frac{9.8 \times 6 \times 6}{2}$

$\Rightarrow h = - 58.8 m$

Let say height above tower till its speed becomes zero is H.

$v^{2} = u^{2} + 2 as$

$\Rightarrow 0 = {(19.6)}^{2} - 2 (9.8) H \Rightarrow H = 19.6 m$

The height from the ground up to which the ball can rise is $h + H = 78.4 m = \frac{392}{5} m$

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