Question

The expression of the trajectory of a projectile is given as $y=px-q{x}^2$ ,
where $y$ and $x$ are respectively the vertical and horizontal displacements and $p$ and $q$ are constants. The time of flight of the projectile is

• A. $\frac{p^2}{4q}$
• B. $\frac{p^2}{2q}$
• C. $\sqrt{\frac{2p}{qg}}$
• D. $p\sqrt{\frac{2}{qg}}$

Answer 1

Answer: (D)

Given, $y=px-q{x}^2$
$y_{\max }$ when $\frac{dy}{dx}=0$
$\Rightarrow p-2qx=0$
$\Rightarrow y_{\max }$ or max height $\left(H\right)=\frac{p^2}{4q}$
Now, $H=y_{max}=\frac{u_y^2}{2g}$
$\Rightarrow \frac{u_y^2}{2g}=\frac{p^2}{4q}$
$\Rightarrow u_y=\sqrt{\frac{g{p}^2}{2q}}$
Also, $T=$ time of flight
$=\frac{2u\sin \theta }{g}=\frac{2u_y}{g}$
$=p\sqrt{\frac{2}{gq}}$