A frog sits at the end of a long board of length L. The board rests on a smooth horizontal surface. The frog wants to jump to the opposite end of the board. What is the minimum take off speed i.e relative to the ground that allows the frog to do the trick? Assume that frog and board have equal masses

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$\sqrt{\frac{gL}{3}}$

$\sqrt{\frac{gL}{1}}$

$\sqrt{\frac{gL}{2}}$

$\sqrt{\frac{gL}{4}}$

answer is C.

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Detailed Solution

$\begin{array}{l} S o \frac{L}{2 u \cos θ} = \frac{2 u \sin θ}{g} \\ \Rightarrow u^{2} = \frac{g L}{2 \sin (2 θ)} \\ \Rightarrow u_{\min = \sqrt{\frac{g L}{2}}} \end{array}$

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