Question

The trajectory of a projectile near the surface of the earth is given as $y = 2x - 9x^2$. If it were launched at an angle $\theta_0$ with speed $\nu_0$ then ($g = 10\, ms^{-2}$) :

• A. $\theta_{0} = \cos^{-1} \left(\frac{1}{\sqrt{5}}\right)$ and $ v_{0} = \frac{5}{3} ms^{-1} $
• B. $\theta_{0} = \sin^{-1} \left(\frac{1}{\sqrt{5}}\right)$ and $ v_{0} = \frac{5}{3} ms^{-1} $
• C. $\theta_{0} = \sin^{-1} \left(\frac{2}{\sqrt{5}}\right)$ and $ v_{0} = \frac{3}{5} ms^{-1} $
• D. $\theta_{0} = \cos^{-1} \left(\frac{2}{\sqrt{5}}\right)$ and $ v_{0} = \frac{3}{5} ms^{-1} $

Answer 1

Answer: (A)

Equation of trajectory is given as $y = 2x - 9x^2 $ ......(1)
Comparing with equation :
$y = x \tan \theta - \frac{g}{2u^2 \cos^2 \theta } . x^2 $ .....(2)
We get;
$\tan \theta = 2$
$\therefore \cos\theta = \frac{1}{\sqrt{5}} $
Also, $ \frac{g}{2u^{2} \cos^{2} \theta} = 9 $
$ \Rightarrow \frac{10}{2\times9 \times \left(\frac{1}{\sqrt{5}}\right)^{2}} = u^{2} $
$ \Rightarrow u^{2} = \frac{25}{9} $
$ \Rightarrow u= \frac{5}{3} m/s $