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. 10 m
• B. 7.5 m
• C. 5 m
• D. 2.5 m

Answer 1

Answer: (C)

5 m

Answer 2

$T = \frac{2u\sin\theta}{g} = 2sec$ (given)

$\therefore u\sin\theta = 10$

Now $H = \frac{u^{2}\sin^{2}\theta}{2g} = \frac{(10)^{2}}{2 \times 10} = 5m.$