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. $\frac{u^{2}}{g}$

Answer 1

Answer: (A)

$\frac{u^{2}}{4g}$

Answer 2

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)
$\therefore u \cos \theta =\frac{u}{\sqrt{2}} \Rightarrow \cos \theta=\frac{1}{\sqrt{2}}$
$\Rightarrow \theta =45^{\circ}$
The maximum height is given by the formula
$H=\frac{u^{2} \sin ^{2} \theta}{2 g}=\frac{u^{2}}{2 g}\left(\frac{1}{\sqrt{2}}\right)^{2}=\frac{u^{2}}{4 g}$