A body is projected with a velocity u so that the horizontal range is twice the maximum height.The maximum height is
(A)u^2/2g
(B)3u^2/2g
(C)2u^2/5g
(D)5u^2/2g
Explain with process
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Option (c)
Angle of project with horizontal = Ф
Range = R = u² sin 2Ф / g
MAximum Height = H = u² Sin² Ф / 2 g
So R = 2 H
=> 2 SinФ Cos Ф = 2 * 1/2 sin² Ф
=> tan Ф = 2 / 1
=> sin Ф = 2/√(1²+2²) = 2 /√5
maximum height = H = 2 u² / (5 g)
Angle of project with horizontal = Ф
Range = R = u² sin 2Ф / g
MAximum Height = H = u² Sin² Ф / 2 g
So R = 2 H
=> 2 SinФ Cos Ф = 2 * 1/2 sin² Ф
=> tan Ф = 2 / 1
=> sin Ф = 2/√(1²+2²) = 2 /√5
maximum height = H = 2 u² / (5 g)
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