Question

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 $(g = 10\, m/s^{2})$

• A. $2.5 \,m$
• B. $5 \,m$
• C. $7.5 \,m$
• D. $10 \,m$

Answer 1

Answer: (B)

$5 \,m$

Answer 2

Since the ball reaches from one player to another in $2\,s$, so the time period of the flite, $T = 2\, s$
$\Rightarrow \frac{2u\, sin\,\theta}{g}=2s$
Here, $u$ is the initial velocity and $\theta$ is the angle of projection.
$\Rightarrow u\,sin\, \theta=g \ldots\left(i\right)$
Now, we know that the maximum height of the projection
$H=\frac{u^{2}\,sin^{2}\,\theta}{2g}$
or $H=\frac{\left(u\,sin\,\theta\right)^{2}}{2g}$
On putting the value of $u\, sin\,\theta$ from E $\left(i\right)$, we have
$H=\frac{g^{2}}{2g}=\frac{g}{2}$
or $H=\frac{g}{2}=\frac{10}{2}\,m$
or $H=5\,m$