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Question

A projectile is projected from ground with velocity v0 at an angle θ with horizontal. What is the velocity of projectile when its vertical displacement is equal to half of the maximum height attained?

A
3v0cosθ
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B
v02sinθ
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C
v02cosθ
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D
5v0
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Solution

The correct option is B v02sinθ
Initial vertical component of the velocity, u=v0sina

Maximum height is given by, v2u2=2gh

At top most height, v=0

Then,

0(v0sina)2=2gh

h=(v0sina)22g

h=h2

h=(v0sina)24g

Vertical component of the velocity at this point

v2u2=2gh

v=(v0sina)22g(v0sina)24g

v=v0sina2

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