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A ball is projected horizontally from the top of a building $19.6 m$ high. If the line joining the point of projection to the point where it hits the ground makes an angle of $45 °$ to the horizontal, the initial velocity of the ball is $({in ms}^{- 1})$ $(g = 9.8 m / s^{2})$ (Round off nearest integer)

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Detailed Solution

$T = \sqrt{\frac{2 H}{g}} = \sqrt{\frac{2 \times 19.6}{9.8}} = 2 \sec \tan 45 ° = \frac{H}{R} R = H R = U x . T 19.6 = U x . (2) U x = 9.8 m / s$

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