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

The correct option is C 5 m

$T=\frac{2usin\theta }{g}=2sec\left(given\right)\therefore usin\theta =10NowH=\frac{u^{2}si{n}^2\theta }{2g}=\frac{(10{)}^2}{2\times 10}=5m$