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# A particle is projected from the ground with an initial velocity of 20 m/s at an angle of 30o with horizontal. The magnitude of change in velocity in a time interval from t=0 to t=0.5s is (g=10m/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 A 5 m/sGiven-u=20m/s , θ=30o , t=0.5sInitial velocity-→u=ucosθ^i+usinθ^j⟹→u=u(cos30o^i+sin30o^j)⟹→u=20(√32^i+12^j)→u=10√3^i+10^jacceleration a=−g^j=−10^j since acceleration is acting downward→v=→u+→at⟹→v=(10√3^i+10^j)−(10^j×0.5)→v=10√3^i+5^jChange in velocity=Δ→v=→v−→u=−5^jMagnitude of change in velocity =5m/sA short method is-Δ→v=acceleration×t=10×0.5=5m/sAnswer-(A)  Suggest Corrections  0      Similar questions  Related Videos   Jumping Off Cliffs
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