A frog sits on the end of a long board of length L = 10 cm. The board rests on a frictionless horizontal table. The frog wants to jump to the opposite end of the board. What is minimum take off speed v in ${\mathrm{ms}}^{-1}$ relative to the ground that the frog follows to do the trick? [Assume that, the board and frog have equal masses.] 

Let the speed of the board be u and frog jumps with angle of inclination to the board $\mathrm{\theta }$, then from law of conservation of momentum in horizontal direction,

              $\mathrm{mv} \mathrm{cos\theta } - \mathrm{mu}=0, \mathrm{u}=\mathrm{v} \mathrm{cos\theta }$                                  …(i)

Let distance moved board be x.

So,                   $L-x=ut$                                                           …(ii)

and                         $x=v \cos \theta t$

Solving above equations, we get

                            $x=\frac{L}{2}$

Also,                    $x = \frac{v^2\sin 2\theta }{g}$

$\Rightarrow$                       $\frac{L}{2}=\frac{v^2\sin 2\theta }{g}$

$\Rightarrow$                       $v=\sqrt{\frac{gL}{2 \sin 2\theta }}$

Hence, v should be minimum for sin 2 $\mathrm{\theta }$ = 1 (i.e. maximum)

$\Rightarrow$                       $v_{\min }=\sqrt{\frac{gL}{2}}=\sqrt{10\times \frac{10}{2}}=\sqrt{50}$     

                                   $=5\sqrt{2} {\mathrm{ms}}^{-1}$