    Question

# A particle is projected from the ground with an initial velocity of 20 m/s at an angle of 30∘ with horizontal. The magnitude of change in velocity in a time interval from t=0 to t=0.5 s is (g=10 m/s2)

A
5 m/s
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B
2.5 m/s
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C
2 m/s
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D
4 m/s
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Solution

## The correct option is D 5 m/sGiven, u=20 m/s, θ=30∘, t=0.5 s →u=ux^i+uy^j →u=20cos30∘^i+20sin30∘^j→u=10√3^i+10^j Let v be the velocity at t=0.5 s⇒v=vx^i+vy^j Horizontal component of velocity in a projectile motion always remains constant ⇒vx=ux=20cos30∘=10√3^i By using first equation of motion, we can say vy=uy−gt vy=20sin30∘−10(0.5)=5 m/s∴v=10√3^i+5^j Change in velocity Δv=v−u=(10√3^i+5^j)−(10√3^i+10^j)Δv=−5 ms−1 |Δv|=5 ms−1  Suggest Corrections  1      Similar questions  Related Videos   Solving n Dimensional Problems
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