Kinematics

JEE Main level, single correct, medium

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Question 221 of 308

A projectile is projected at 30∘30^{\circ} from horizontal with initial velocity 40ms−140m{s}^{-1}. The velocity of the projectile at t=2st=2s from the start will be:

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Answer: (D) 203ms−120\sqrt{3}m{s}^{-1}

It is given that v=40ms−1v=40m{s}^{-1}.
The components of the velocity will be vy=vsin⁡30∘=40ms−12=20ms−1vx=vcos⁡30∘=4032ms−1=203ms−1{v}_{y}=v\sin 30^{\circ}=\dfrac{40m{s}^{-1}}{2}=20m{s}^{-1} {v}_{x}=v\cos 30^{\circ}=\dfrac{40\sqrt{3}}{2}m{s}^{-1}=20\sqrt{3}m{s}^{-1}
The time taken to reach maximum height is T=usin⁡30∘g=20ms−110ms−2=2sT=\dfrac{u\sin 30^{\circ}}{g}=\dfrac{20m{s}^{-1}}{10m{s}^{-2}}=2s
Hence, at T=2s,asvy=0ms−1thereforevnet=vx=203ms−1T=2s,\text{as}{v}_{y}=0m{s}^{-1}\text{therefore}{v}_{net}={v}_{x}=20\sqrt{3}m{s}^{-1}.

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