Question

A ball is thrown from a point at a distance $40$m from a wall of height $15$m. It just clears the wall and then attains maximum height. Find the maximum height if the angle of projection is ${45}^o$.

• A. 60m
• B. 45 m
• C. 16 m
• D. 50 m

Answer 1

The correct option is C 16 m

Given : $x=40m$ $y=15m$

Using equation of trajectory $y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$

where $\theta ={45}^o$ $⟹\tan 45=1$ $cos45=\frac{1}{\sqrt{2}}$

Or $15=40\times 1-\frac{10\times {40}^2}{2\times u^2\times \frac{1}{2}}$

Or $\frac{10\times {40}^2}{u^2}=25$

$⟹u^2=640$

Maximum height attained $H_{max}=\frac{u^2{\sin }^2\theta }{2g}$

$⟹H_{max}=\frac{640\times \frac{1}{2}}{2\times 10}=16m$