Verified Solution • 2025–26 Session moderate

QJEEM22Phee40: JEE Main 2022 Physics - Class 11 JEEM22Phee40 Solution: A projectile is projected with velocity of 25 m/s at an angle…

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

[use g = 10 m/s2]

Multiple Choice Options

  • A 12sin⁡−1(5t24R){1 \over 2}{\sin ^{ - 1}}\left( {{{5{t^2}} \over {4R}}} \right)
  • B 12sin⁡−1(4R5t2){1 \over 2}{\sin ^{ - 1}}\left( {{{4R} \over {5{t^2}}}} \right)
  • C tan⁡−1(4t25R){\tan ^{ - 1}}\left( {{{4{t^2}} \over {5R}}} \right)
  • D cot⁡−1(R20t2){\cot ^{ - 1}}\left( {{R \over {20{t^2}}}} \right)

Step-by-Step Verified Solution

JEE Main 2022 (Online) 24th June Morning Shift Physics - Motion in a Plane Question 37 English Explanation

t=25sin⁡θgt = {{25\sin \theta } \over g}

and, R=(25)2(2sin⁡θcos⁡θ)gR = {{{{(25)}^2}(2\sin \theta \cos \theta )} \over g}

⇒R=25×25×2g×gt25×cos⁡θ\Rightarrow R = {{25 \times 25 \times 2} \over g} \times {{gt} \over {25}} \times \cos \theta

⇒R=50tcos⁡θ\Rightarrow R = 50t\cos \theta

∴\therefore tanθ=gt25×50tRtan\theta = {{gt} \over {25}} \times {{50t} \over R}

=20t2R= {{20{t^2}} \over R}

⇒θ=cot⁡−1(R20t2)\Rightarrow \theta = {\cot ^{ - 1}}\left( {{R \over {20{t^2}}}} \right)

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