Question

A ball projected at an angle ${45}^0$ with velocity 20 $\sqrt{2}m{s}^{-1}$ just clears a wall which is 60 m away. Find the height of the wall.

• A. 10 m
• B. 15 m
• C. 20 m
• D. 25 m

Answer 1

The correct option is B 15 m

$ucos\theta t=60\Rightarrow 20\sqrt{2}cos{45}^{\circ }\times t=60\Rightarrow t=3sNowh=usin\theta -\frac{1}{2}g{t}^2=\left(20\sqrt{2}sin{45}^{\circ }\right)\times 3-\frac{1}{2}\times 10\times 3^2=15m$