A particle is projected from the bottom of an inclined plane of angle 30∘ with a velocity of 30m/s at an angle of 60∘ with the horizontal. Find the time of flight.
(Take g=10m/s2)
A
√3s
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B
1√3s
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C
2√3s
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D
2s
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Solution
The correct option is C2√3s We know that time of flight of a projectile on an inclined plane is given by T=2usin(α−θ)gcosθ
Here α=60∘ and θ=30∘ ∴T=2×30×sin(60∘−30∘)gcos30∘ ⇒T=2√3s