JEE Main 2022PhysicsMotion In Two DimensionsEasyMCQ

JEE Main 2022 — Motion In Two Dimensions Question with Solution

JEE Main 2022 (24 Jun Shift 1)

Question

A projectile is projected with velocity of 25 m s-1 at an angle θ with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of θ will be : [use use g=10 m s-2]

Choose an option

Show full solutionCorrect option: D
Correct answer
Dcot-1R20t2

Step-by-step explanation

From the information given in the question, after time t the inclination of the projectile becomes zero i.e. the projectile is at the highest point. Therefore,

t=T2=usinθg.

The rang of the projectile R=2u2sinθcosθg.

Now,

Rt2=2u2sinθcosθg×g2u2sin2θ=2gcotθ⇒θ=cot-1R20t2

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About this question

This is a previous-year question from JEE Main 2022, covering the Motion In Two Dimensions chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.