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Q.
A ball thrown by one player reaches the other in $2\, s$. The maximum height attained by the ball above the point of projection will be $(g=10\,m/s^{2})$
AFMCAFMC 2012
Solution:
Since, the ball reaches from one player to another in $2 s$, so the time period of the flight, $T=2 s$
$\Rightarrow \frac{2 u \sin \theta}{g}=2 s$
Here, $u$ is the initial velocity and $\theta$ is the angle of projection.
$\Rightarrow u \sin \theta=g$
Now, we know that the maximum height of the projection
$H=\frac{u^{2} \sin ^{2} \theta}{2 g}$
or $H=\frac{(u \sin \theta)^{2}}{2 g}$
On putting the value of $u \sin 0$ from Eq. (i), we have
$H = \frac{g^2}{2g} = \frac{g}{2}$
or $H=\frac{g}{2}=\frac{10}{2} m$
or $H=5\, m$
