A ball is thrown vertically upwards with a velocity of $19.6 m s^{- 1}$ from the top of a tower. The ball strikes the ground after $8 s$ . The height from the ground up to which the ball can rise will be $(\frac{k}{5}) m$ . The value of $k$ is__________ $(u s e g = 9.8 m / s^{2})$

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answer is 882.

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Detailed Solution

$t_{a} = \frac{u}{g} = \frac{19.6}{9.8} = 2 \sec t_{d} = 8 - 2 = \sqrt{\frac{2 h m g}{g}} = h \max = \frac{36 \times 9.8}{2} = 176.4 = \frac{k}{5} k = 882$

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