Explanation

(b) Since the ball reaches from one player to another in 2 s, so the time period of the flite, $T = 2s$
$or\ \ \frac{2usin\theta}{g} = 2$
$or\ \ usin\theta = g\ \ ..(i)$
Now, we know that the maximum height of the projection.
$H = \frac{u^{2}\sin^{2}\theta}{2g}$
$or\ \ H = \frac{(usin\theta)^{2}}{2g}$
On putting the value of$\ usin\theta\ $from Eq.(i), we have$\ $
$H = \frac{g^{2}}{2g} = \frac{g}{2}$
$or\ \ H = \frac{g}{2} = \frac{10}{2}m\ or\ H = 5m$