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le is thrown with an initial velocity ofv.i+ v.j) ms-1. If the range of the projectle is2 A projectidouble the maximum height, then v,(1) 2) 2, (3) 3V (4) 4v, |
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Answer» Range = Vx.T and T = 2Vy/g so, Range = 2Vx.Vy/g and maximum height = H = Vy²/2g now, R = 2H => 2Vx.Vy/g = 2*(Vy²/2g)=> Vx.Vy = 1/2*(Vy.Vy)=> Vy = 2Vx option 2 |
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