Question

A particle is projected with a velocity u, at an angle α, with the horizontal. Time at which its vertical component of velocity becomes half of its net speed at the highest point will be

A
u2g
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B
u2g(sinαcosα)
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C
u2g(2cosαsinα)
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D
u2g(2sinαcosα)
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Solution

The correct option is D u2g(2sinα−cosα)We know that for the projectile motion with initial velocity u thrown at an angle α with horizontal, net speed at the highest point will be only along horizontal direction and its value will be ucosα Hence, half of speed at the highest point =ucosα2. Time taken for the vertical component of velocity to be half of the net speed at highest point can be found by the first equation of motion, vy=uy+ayt, where ay=−g, taking downward direction as negative, uy=usinα ∴ucosα2=usinα−gt t=u2g(2sinα−cosα)

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