Question

A frog sits on the end of a ball board of length $L=5m$. The board rests on a frictionless horizontal table. The frog wants to jump to the opposite end of the board. What is the minimum take-off speed (in m/s), i.e., relative to ground 'v' that allows the frog to do the trick? The board and the frog have equal masses

Answer 1

Let $\theta$ be the angle with horizontal at which it jumps and p be the momentum it delivers.
given the mass of both is same. so ,
Horizontal momentum on ball board=$p\cos \left(\theta \right)$
$\Rightarrow velocit{y}_{horizontal}=\frac{p\cos \left(\theta \right)}{m}$
velocity of frog$=\frac{p}{m}$
horizontal component of velocity of frog$=\frac{p}{m}\cos \left(\theta \right)$
And vertical velocity$=\frac{p}{m}\sin \left(\theta \right)$
therefore relative velocity$=\frac{2p\cos \left(\theta \right)}{m}$
Time of flight of frog$=\frac{\frac{p\sin \left(\theta \right)}{m}}{2g}$
$\Rightarrow \frac{2p\cos \left(\theta \right)}{m}\frac{\frac{p\sin \left(\theta \right)}{m}}{2g}=5$
since $p=mv$
by substituting,
$v=\frac{10}{\sqrt{\sin \left(2\theta \right)}}$
minimum value of $v=10m/s$