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2015-09-18T20:07:09+05:30

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Max h=(v₀)²sin²(q) /2 g

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MASelvam IITian
Booksworm.
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2015-09-18T20:27:47+05:30

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U = initial velocity of the projectile
θ = angle of projection with the horizontal
H = maximum height above the horizon that the projectile attains
   = u² Sin²θ / (2 g)

derivation:
   initial velocity in vertical direction = u sinθ
   v² = u²  - 2 g H
   v = 0  when the projectile reaches the top.
      distance travelled  H = u² Sin² θ / 2 g

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To select best answer two people must answer.So, please wait for me to come through my next account,Studious.I;m sorry, I won't choose yours as best for the reason I said befor.