JEE Main 2022 · 24 Jun · Shift 1 · Physics Q35

← JEE Main 2022 · 24 Jun · Shift 1

PhysicsSingle correct+4 / -1Question 35 of 90

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 $g=10\ \mathrm{m/s^2}$ ]

\dfrac12\sin^{-1}\left(\dfrac{5t^2}{4R}\right)

\dfrac12\sin^{-1}\left(\dfrac{4R}{5t^2}\right)

\tan^{-1}\left(\dfrac{4t^2}{5R}\right)

\cot^{-1}\left(\dfrac R{20t^2}\right)

Answer and worked solution

Official answer: D

Time to the top. Inclination is zero at the highest point: $t=\dfrac{u\sin\theta}g$ , so $\sin\theta=\dfrac{10t}{25}=\dfrac{2t}5$ .

Range. $R=\dfrac{2u^2\sin\theta\cos\theta}g=\dfrac{2(625)}{10}\cdot\dfrac{2t}5\cos\theta=50t\cos\theta$ , so $\cos\theta=\dfrac R{50t}$ .

Angle. $\tan\theta=\dfrac{2t/5}{R/(50t)}=\dfrac{20t^2}R$ , so $\theta=\cot^{-1}\left(\dfrac R{20t^2}\right)$ .

(D)

Topics:Projectile motion

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JEE Main 2022 · 24 Jun · Shift 1, Physics question 35