A projectile is thrown so as to have a maximum possible horizontal range of 400m. 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=1⇒2θ=90∘⇒θ=45∘ According to question R=u2g=400m We know that velocity is minimum at the highest point of projectile (i.e. H) by using the relation. Rtanθ=4H 400tan45∘=4H ⇒H=100
∴ Coordinates of the point where velocity is minimum is (200,100)