A ball is projected horizontally with a speed V from the top of a plane inclined at angle 450 with the horizontal. How far from the point of projection will the ball strike the plane?

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V2g

2V2g

22V2g

answer is D.

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Detailed Solution

down the plane Range =R=u2gcos2⁡β[sin⁡(2α+β)+sin⁡β]here α=0=angle of projection is along horizontal line;substituteR=u2gcos2⁡45∘sin⁡45∘+sin⁡45∘R=u2g12×2×12R=22V2g [ Here u=v]

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