Question

A ball thrown by one player reaches the other in $2$ sec. The maximum height attained by the ball above the point of projection will be about :

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

Answer 1

Answer: (C)

$5m$

Given,

$T=2sec$

$g=10m/s^2$

From the projectile motion,

The time of flight is given by

$T=\frac{2usin\theta }{g}=2sec$

$usin\theta =g$

Squaring both side, we get

$u^{2}si{n}^2\theta =g^2$. . . . . . . .(1)

The maximum height attained by the ball is given by

$H=\frac{u^{2}si{n}^2\theta }{2g}$. . . . . .(2)

Substitute equation (2) in equation (1), we get

$H=\frac{g^2}{2g}$

$H=\frac{g}{2}=\frac{10}{2}=5m$

The correct option is C.