Question

A body is thrown up with a speed u, at an angle of projection $\theta$. If the speed of the projectile becomes $\frac{u}{\sqrt{2}}$ on reaching the maximum height, then the maximum vertical height attained by the projectile is

• A. $\frac{u^2}{4g}$
• B. $\frac{u^2}{3g}$
• C. $\frac{u^2}{2g}$
• D. $\frac{u^2}{g}$
• E. $\frac{2u^2}{g}$

Answer 1

The correct option is A $\frac{u^2}{4g}$

Speed of projectile at maximum height
$=\frac{u}{\sqrt{2}}$ .........(i)
Also, we know that speed of a projectile at maximum height $=u\cos \theta$ ...........(ii)
$\therefore u\cos \theta =\frac{u}{\sqrt{2}}\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}$
$\Rightarrow \theta ={45}^o$
The maximu height is given by the formula
$H=\frac{u^2{\sin }^2\theta }{2g}=\frac{u^2}{2g}{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{u^2}{4g}$.