A particle is projected with a velocity u at an angle alpha with the horizontal. At what time its vertical component of velocity is half of its net speed at heighest point

Question

Khimraj · Answer

net speed at highest point = ucos then v = u -gt ucos /2 = usin – gt Then t = usin /g – ucos /2g Hope it clears.

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A particle is projected with a velocity u at an angle alpha with the horizontal. At what time its vertical component of velocity is half of its net speed at heighest point

A particle is projected with a velocity u at an angle alpha with the horizontal. At what time its vertical component of velocity is half of its net speed at heighest point

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A particle is projected with a velocity u at an angle alpha with the horizontal. At what time its vertical component of velocity is half of its net speed at heighest point

A particle is projected with a velocity u at an angle alpha with the horizontal. At what time its vertical component of velocity is half of its net speed at heighest point

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