Question

A ball thrown by one player reaches the other in $2s$. The maximum height attained by the ball above the point of projection will be $(g=10m/s^2)$

• A. 2.5 m
• B. 5 m
• C. 7.5 m
• D. 10 m

Answer 1

Answer: (B)

Since, the ball reaches from one player to another in $2s$, so the time period of the flight, $T=2s$
$\Rightarrow \frac{2u\sin \theta }{g}=2s$
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
$H=\frac{u^2{\sin }^2\theta }{2g}$
or $H=\frac{(u\sin \theta {)}^2}{2g}$
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 $H=5m$