Physics · 2019
JEE Main · 12 April 2019, Shift 1 · Q2
The trajectory of a projectile near the surface of the earth is given as y=2 x-9 x^2. If it were launched at an angle θ_0 with speed v_0 then ( g =10…
The trajectory of a projectile near the surface of the earth is given as $\displaystyle y=2 x-9 x^2$. If it were launched at an angle $\displaystyle \theta_0$ with speed $\displaystyle v_0$ then $\displaystyle \left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$ :
Official answer
(2)
$\displaystyle \theta_0=\cos ^{-1}\left(\frac{1}{\sqrt{5}}\right) \text { and } v_0=\frac{5}{3} \mathrm{~ms}^{-1}$
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.