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Question

# A ball is projected horizontally with a speed v from the top of a plane inclined at an angle 45∘ with the horizontal. How far from the point of projection will the ball strike the plane?

A
v2g
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B
2v2g
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C
2v2g
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D
2[2v2g]
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Solution

## The correct option is D √2[2v2g] Let l be the distance along inclined plane at which the ball will strike. Let the horizontal distance covered by the ball before striking be x. ⇒x=vt (i) Vertical distance covered by the ball is given by y=12gt2 (ii) Using value of t from (i) in (ii), we get y=gx22v2 (iii) From the figure, tan45∘=yx or y=x ⇒x=gx22v2 (from (iii)) ⇒x=2v2g So, y=x=2v2g From the figure, we get, l2=x2+y2=√2[2v2g] is the distance along the inclined plane at which the ball will hit. Or Alternative Method 1: Applying second eqn of motion, in the direction perpendicular to the inclined plane, s=0 i.e vcos450t−12gsin450t2=0 ⇒t=2vg Hence distance along the inclined plane x=vsin45∘t+12gsin45∘t2=v√2×2vg+12×g√2×4v2g=2√2v2g Alternative Method 2 : Since the ball is projected horizontally, horizontal distance covered x=v√2yg and x=y as angle is 45∘⇒√x=v√2g⇒x=2v2g=y∴AB=x√2=2√2v2g

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