Question

The path of projectile is represented by: $y=Px-Q{x}^2$

Answer 1

The path of particle is $y=Px-Q{x}^2$ ............(1)

Range : $y=0$ at $x=\left(x_R\right)$

$\therefore$ $0=P\left(x_R\right)-Q\left(x_R^2\right)$ $⟹$ Range $x_R=\frac{P}{Q}$

Maximum height : $\frac{dy}{dx}{|}_{x=x_1}=0$

$\therefore$ $P-2Q\left(x_1\right)=0$ $⟹x_1=\frac{P}{2Q}$

Thus maximum height $H_{max}=y{|}_{x=x_1}=P\times \frac{P}{2Q}-Q\times \frac{P^2}{4Q^2}=\frac{P^2}{4Q}$

Angle of projection :

Differentiating (1) w.r.t time (t), we get $u_y=P{u}_x-\left(2Qx\right)u_x$ ........(2)

Put $x=0$ in (2), we get $u_y=P{u}_x$

$\therefore$ $tan\theta =\frac{u_y}{u_x}=P$

Time of flight :

Using $x_R=u_{x}T$

$⟹$ $u_x=\frac{x_R}{T}=\frac{P}{QT}$

$S_y=0$ for $t=T$ (time of flight)

Using $S_y=u_{y}t-\frac{1}{2}g{t}^2$

$\therefore$ $0=u_{y}T-\frac{g{T}^2}{2}$ $⟹T=\frac{2u_y}{g}=\frac{2u_{x}P}{g}$

$\therefore$ $T=\frac{2P}{g}\times \frac{P}{QT}$ $⟹T=\sqrt{\frac{2}{Qg}}$ $P$