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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 vcos450t12gsin450t2=0
t=2vg

Hence distance along the inclined plane
x=vsin45t+12gsin45t2=v2×2vg+12×g2×4v2g=22v2g

Alternative Method 2 :

Since the ball is projected horizontally, horizontal distance covered
x=v2yg and x=y as angle is 45x=v2gx=2v2g=yAB=x2=22v2g

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