A projectile is projected with velocity of 25 m/s at an angle θ with… | JEE Main 2022 (24 Jun, Shift 1)

Q23JEE Main 2022 · 24 JUN · SHIFT 1Medium

Time of Flight, Maximum Height & Range

A projectile is projected with velocity of 25 m/s at an angle $\theta$ with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of $\theta$ will be:

[use g = 10 m/s2]

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Answer: (d)

Inclination zero means the top: $t = \frac{u\sin\theta}{g} \Rightarrow u\sin\theta = 10t$
$R = \frac{2(u\sin\theta)(u\cos\theta)}{g} = 2t\,u\cos\theta \Rightarrow u\cos\theta = \frac{R}{2t}$
$\tan\theta = \frac{10t}{R/2t} = \frac{20t^2}{R} \Rightarrow \cot\theta = \frac{R}{20t^2}$
So the answer is (d) $\cot^{-1}\left(\frac{R}{20t^2}\right)$ .

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