1take the vertical componenet of velocity vsin∂ and distance as v2sin2∂ /4g and apply the 2nd equation of motion
how much time does a projectile take to reach half of its greatest height??
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7 Answers
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1\frac{v^2 sin^2\alpha }{4g}= vsin\alpha .t + \frac{1}{2} gt^2
solve dis for t
49@जय: Don't u think that in that equation u have used erronous sign convention?
cos i thnk
initially te vel component is vsin@ in upward direction whilst the accn is g in downward direction
and so if H and Vsin@ are entertained wid a + sign, g must be made to be devoid of that privilege :P
62you would have been better off using a graph of velocity time graph..
1/2 (at) . t = 2 . 1/2 (t-x)a(t-x)
so t/(t-x) = √2
x=(1-1/√2)t