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A man is riding on a horse. He is trying to jump the gap between two symmetrical ramps of snow separated by a distance W as shown in figure. He launches off the first ramp with a speed V_L. The man and the horse have a total mass m. and their size is small as compared to W The value of initial launch speed V_L which will put the horse exactly at the peak of the second ramp is

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$\sqrt{\frac{W g}{\sin (θ / 2) \times \cos (θ / 2)}}$

$\sqrt{\frac{W g}{\sin θ \times \cos θ}}$

$\sqrt{\frac{W g}{2 \sin θ \cos θ}}$

$\sqrt{\frac{2 W g}{\sin θ \cos θ}}$

answer is C.

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Detailed Solution

Due to symmetry, the maximum of the trajectory lies halfway across the gap.

Therefore, range of projectile motion = W

$\begin{array}{l} \Rightarrow W = \frac{2 V_{L}^{2} \sin θcos θ}{g} \\ ∴ VL = \sqrt{\frac{Wg}{2 \sin θcos θ}} \end{array}$

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