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A ball is thrown from the ground level so as to clear a wall 3m high, located at a distance 4m, and falls to the other side at a distance of 16m from the wall. Find the initial velocity (in m/s) of the ball. $(g = 10 m / s^{2})$

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answer is 14.16.

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Detailed Solution

Range, $R = 4 + 16 = 20 m$

$y = (\tan α) x - \frac{x^{2} \tan α}{R}$

$3 = (\tan α) 4 - \frac{16 \tan α}{20}$

$3 = \tan α (4 - \frac{4}{5})$

$\tan α = \frac{15}{16}$

$R = \frac{u^{2} \sin 2 α}{g} \Rightarrow 20 = \frac{u^{2} \times 2 \times (\frac{15}{\sqrt{481}}) (\frac{16}{\sqrt{481}})}{10}$

$\Rightarrow u^{2} = 200.4 \Rightarrow u = 14.16 m s^{- 1}$

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