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(C) The time taken to reach the maximum height is $t_a = \frac{u}{g} = \frac{19.6}{9.8} = 2 \, s$.The maximum height reached by the ball from the poin

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$392$

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(C) The time taken to reach the maximum height is $t_a = \frac{u}{g} = \frac{19.6}{9.8} = 2 \, s$.The maximum height reached by the ball from the point of projection is $H = \frac{u^2}{2g} = \frac{(19.6)^2}{2 	imes 9.8} = \frac{19.6 	imes 19.6}{19.6} = 19.6 \, m$.Let $h$ be the height of the tower. Using the equation of motion $s = ut + \frac{1}{2}at^2$ for the total displacement from the top of the tower to the ground:$-h = (19.6)(6) + \frac{1}{2}(-9.8)(6)^2$$-h = 117.6 - 4.9 	imes 36$$-h = 117.6 - 176.4 = -58.8 \, m$,so $h = 58.8 \, m$.The maximum height from the ground is $H_{max} = h + H = 58.8 + 19.6 = 78.4 \, m$.Given $H_{max} = \frac{k}{5} = 78.4$,we get $k = 78.4 	imes 5 = 392$.

$A$ ball is thrown vertically upwards with a velocity of $19.6 \, m/s$ from the top of a tower. The ball strikes the ground after $6 \, s$. The height from the ground up to which the ball can rise is given by $\left(\frac{k}{5}\right) \, m$. The value of $k$ is ..... (use $g = 9.8 \, m/s^2$)

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