Question

If a particle is projected from origin and it follows the trajectory $y=x-\frac{1}{4}{x}^2$, then the time of flight is ($g$ = acceleration due to gravity)

• A. $\frac{\sqrt{2}}{\sqrt{g}}$
• B. $\frac{2\sqrt{2}}{\sqrt{g}}$
• C. $\frac{2}{\sqrt{g}}$
• D. $\frac{3\sqrt{2}}{\sqrt{g}}$

Answer 1

The correct option is B $\frac{2\sqrt{2}}{\sqrt{g}}$

Equation of trajectory, $y=x\tan \theta -\frac{1}{2}\frac{g{x}^2}{u^2{\cos }^2\theta }$
Comparing with
$y=x-\frac{1}{4}{x}^2$
$\tan \theta =1\dots \dots \left(i\right)$
$u\cos \theta =\sqrt{2g}\dots \dots \left(ii\right)$
From (i) and (ii)
$u\sin \theta =\sqrt{2g}$
$T=\frac{2u\sin \theta }{g}=\frac{2\sqrt{2g}}{g}=\frac{2\sqrt{2}}{\sqrt{g}}$
$\therefore$ (b)