Question

The path of projectile is represented by $y=Px-{Qx}^2$.

Answer 1

R.E.F image

$y=Px-Q{x}^2$

we know that,

$y=x\tan \theta -\frac{g{x}^2}{2U^{2}co{s}^2\theta }$

So comparing,

$P=tan\theta$

$Q=\frac{q}{2U^{2}co{s}^2\theta }$

$cos\theta =\frac{1}{\sqrt{P^2+1}}$

$sin\theta =\frac{P}{\sqrt{P^2+1}}$

$U^2=\frac{g(P^2+1)}{2\theta }$

(i) Range $=\frac{2U^{2}sin\theta cos\theta }{g}$

$=2\times g\frac{(P^2+1)}{2\theta \times g}\times \frac{P}{(P^2+1)}$

$\Rightarrow P/Q$

(iii) max height $=\frac{U^{2}si{n}^2\theta }{2g}$

$=\frac{g(P^2+1)}{2\theta \times 2g}\times \frac{P^2}{(P^2+1)}$

$=P^2/4\theta$

(iii) Time of flight $=\frac{R}{Ucos\theta }$

$=\frac{P/Q}{\sqrt{g\frac{(P^2+1)}{2Q}}}\times \frac{1}{\sqrt{P^2+1}}$

$=\sqrt{\frac{2}{Qg}}P$

(iv) $tan\theta =P$