JEE Main 2019PhysicsMotion In Two DimensionsHardMCQ

JEE Main 2019 — Motion In Two Dimensions Question with Solution

JEE Main 2019 (12 Apr Shift 1)

Question

The trajectory of a projectile near the surface of the earth is given as y=2x-9x2. If it were launched at an angle θ0 with speed v0 then g=10 m s-2:

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Show full solutionCorrect option: A
Correct answer
Aθ0=cos-1⁡15 and v0=53 ms-1

Step-by-step explanation

y=2x-9x2
Comparing above equation with general equation of trajectory of projectile y=xtan⁡θ1-xR
tan θ=2
⇒sin θ=25 or cos θ=15
θ=sin-125 or θ=cos-115
And R = 2/9
R=v02sin⁡2θg=29
After substituting the value for g and sin 2θ. We get v0=35 m/s

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About this question

This is a previous-year question from JEE Main 2019, 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.