JEE PYQ: Motion In Two Dimensions Question 17
Question 17 - 2019 (12 Apr Shift 1)
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 ($g = 10$ ms$^{-2}$):
(a) $\theta_0 = \sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$ and $v_0 = \frac{5}{3}$ ms$^{-1}$ (b) $\theta_0 = \cos^{-1}\left(\frac{2}{\sqrt{5}}\right)$ and $v_0 = \frac{3}{5}$ ms$^{-1}$ (c) $\theta_0 = \cos^{-1}\left(\frac{1}{\sqrt{5}}\right)$ and $v_0 = \frac{5}{3}$ ms$^{-1}$ (d) $\theta_0 = \sin^{-1}\left(\frac{2}{\sqrt{5}}\right)$ and $v_0 = \frac{3}{5}$ ms$^{-1}$
Show Answer
Answer: (c)
Solution
$\tan\theta = 2$, so $\cos\theta = 1/\sqrt{5}$. From $g/(2u^2\cos^2\theta) = 9$: $u = 5/3$ m/s.
BETA
