Q 11-03-167JEE MainJEE Main 2019 (12 Apr, Shift 1)Medium
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 $v_0$ then ($g = 10\ \text{m s}^{-2}$):
Answer: (A) $\theta_0 = \cos^{-1}\dfrac{1}{\sqrt5}$ and $v_0 = \dfrac53\ \text{m s}^{-1}$
Compare with $y = x\tan\theta_0 - \dfrac{gx^2}{2v_0^2\cos^2\theta_0}$:
$\tan\theta_0 = 2$, so $\cos\theta_0 = \dfrac{1}{\sqrt5}$.
$$\frac{10}{2v_0^2\cos^2\theta_0} = 9 \Rightarrow v_0^2\cdot\frac15 = \frac59 \Rightarrow v_0 = \frac53\ \text{m/s}$$
Solution by Sreeraj P, M.Sc Physics