Question

the expression of the trajectory of the particle 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. $p\sqrt{\frac{2p}{qg}}$

Answer 1

The correct option is D $p\sqrt{\frac{2p}{qg}}$

Equation of trajectory in projectile motion is given in question is,

$y=px-q{x}^2$. . . . . . . . . .(1)

For maximum height, $\frac{dy}{dx}=0$

$p-2qx=0$

$x=\frac{p}{2q}$

Substitute the value of $x$ is

$Y_{max}=H=p\frac{p}{2q}-q(\frac{p}{2q}{)}^2$

$H=\frac{p^2}{4q}$. . . . . . . . .(2)

The maximum height in the projectile motion,

$H=\frac{u^{2}si{n}^2\theta }{2g}$. . . . . . . (3)

where, $u=$ initial velocity of the projectile

$\theta =$ angle of projection

From equation (2) and (3), we get

$\frac{p^2}{4q}=\frac{u^{2}si{n}^2\theta }{2g}$

$usin\theta =\sqrt{\frac{p^{2}g}{2q}}$

Time of flight in projectile motion,

$T=\frac{2usin\theta }{g}$

$T=\frac{2}{g}\sqrt{\frac{p^{2}g}{2q}}$

$T=\sqrt{\frac{2p^2}{qg}}$