Question

The expression of the trajectory of a projectile is given as $ y = px - qx^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 - qx^{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^{2}_{y}}{2g} $
$ \Rightarrow \frac{u_{y}^{2}}{2g} = \frac{p^{2}}{4q} $
$ \Rightarrow u_{y} = \sqrt{\frac{gp^{2}}{2q}} $
Also, $T =$ time of flight
$= \frac{2u\, \sin\, \theta}{g} = \frac{2u_{y}}{g} $
$ = p\sqrt{\frac{2}{gq}} $