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Question

A projectile is thrown so as to have a maximum possible horizontal range of 400 m. Taking the point of projection as the origin, the co-ordinates of the point where the velocity of the projectile is minimum is

A
(400,100)
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B
(200,100)
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C
(400,200)
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D
(200,200)
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Solution

The correct option is B (200,100)
For R to be maximum sin2θ should be maximum
sin2θ=12θ=90θ=45
According to question
R=u2g=400 m
We know that velocity is minimum at the highest point of projectile (i.e. H)
by using the relation.
Rtanθ=4H
400 tan45=4H
H=100

Coordinates of the point where velocity is minimum is (200,100)

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