Question

The trajectory of a projectile near the surface of the earth is given as $y=2x-9x^2$. Then choose the correct option(s): [Take $g=10{\text{m/s}}^2$]

• A. Angle of projection is ${\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)$
• B. Angle of projection is ${\cos }^{-1}\left(\frac{1}{\sqrt{5}}\right)$
• C. Speed of projection is $\frac{3}{5}\text{m/s}$
• D. Speed of projection is $\frac{5}{3}\text{m/s}$

Answer 1

Answer: (B), (D)

The correct options are
B Angle of projection is ${\cos }^{-1}\left(\frac{1}{\sqrt{5}}\right)$
D Speed of projection is $\frac{5}{3}\text{m/s}$

Equation of Trajectory for projectile is,
$y=2x-9x^2...\left(i\right)$
The standard equation of trajectory for a projectile,
$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }...\left(ii\right)$
On comparing Eq.$\left(i\right)$ and $\left(ii\right)$,
$\tan \theta =2...\left(iii\right)$
& $\frac{g}{2u^2{\cos }^2\theta }=9...\left(iv\right)$


From Eq.$\left(iii\right)$
$\tan \theta =2=\frac{2}{1}$
$\Rightarrow \cos \theta =\frac{1}{\sqrt{5}}$ [From diagram]
$\Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{\sqrt{5}}\right)$
Similarly,
$\sin \theta =\frac{2}{\sqrt{5}}$ [From diagram]
$\Rightarrow \theta ={\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)$
From Eq.$\left(iv\right)$,
$\frac{10}{2\times u^2\times \left(\frac{1}{5}\right)}=9$
$\Rightarrow u=\sqrt{\frac{5\times 5}{9}}$
$\therefore u=\frac{5}{3}\text{m/s}$
Here $\theta$ is the angle of projection and $u$ is the initial speed of projectile.