Question

The path of projectile is represented by $y=Px-Q{x}^2$
$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{(a)}& \text{Range}& \text{(p)}& \frac{P}{Q}\\ \text{(b)}& \text{Maximum height}& \text{(q)}& P\\ \text{(c)}& \text{Time of flight}& \text{(r)}& \frac{P^2}{4Q}\\ \text{(d)}& \text{Tangent of angle of projection}& \text{(s)}& \sqrt{\frac{2}{Qg}}P\end{array}$

• A. $a\to s,b\to q,c\to r,d\to q$
• B. $a\to p,b\to r,c\to q,d\to q$
• C. $a\to p,b\to r,c\to s,d\to q$
• D. $a\to p,b\to q,c\to r,d\to s$

Answer 1

The correct option is C $a\to p,b\to r,c\to s,d\to q$

$a\to pb\to rc\to sd\to q$
$y=Px-Q{x}^2$ (or) (a)
We know path of trajectory of a projectile is,
$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }\left(b\right)$
Comparing (a) and (b),
$p=\tan \theta \left(1\right)$
and $Q=\frac{g}{2u^2{\cos }^2\theta }\left(2\right)$
$\therefore Q=\frac{g}{2u^2{\cos }^2\theta }\times \frac{\sin \theta }{\sin \theta }$
$=\frac{g}{2u^2\cos \theta .\sin \theta }.\tan \theta$
$\Rightarrow Q=\frac{\tan \theta }{\left(\frac{u^2\sin 2\theta }{g}\right)}\left(3\right)$
$\therefore Q=\left(\frac{\tan \theta }{R}\right)$
$\Rightarrow Q=\frac{P}{R}$

a) $Q=\frac{\tan \theta }{R}\therefore R=\frac{P}{Q}$

b) $H=\frac{u^2{\sin }^2\theta }{2g}\times \frac{\cos \theta }{\cos \theta }\times \frac{2}{2}$
$=\frac{u^2(2\sin \theta .\cos \theta )}{4g}\times \frac{\sin \theta }{\cos \theta }$
$=\frac{u^2\left(\sin 2\theta \right)}{4g}\times \tan \theta$
$\Rightarrow H=\frac{P^2}{4Q}$

c) $T=\frac{2u\sin \theta }{g}$
$\therefore T^2=\frac{4u^2{\sin }^2\theta }{g^2}\times \frac{2}{2}$
$=\frac{4}{g}\left(\frac{u^2{\sin }^2\theta }{2g}\right)\times 2$
$=\frac{8}{g}\times H=\frac{8}{g}\times \frac{P^2}{4Q}=\frac{2P^2}{gQ}$
$\Rightarrow T=\left(\sqrt{\frac{2}{Qg}}\right)P$

d) $\tan \theta =P$