A body is projected with velocity of 40m/s after 2s it crosses a vertical pole of height 20.4m calc ulate the angle of projection and horizontal range.
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u = 40 m/s. angle of projection = Ф. g = 9.8 m/s².
Given at t = 2 sec, y = 20.4 m,
y = u SinФ t - 1/2 g t²
20.4 = 80 SinФ - 1/2 * 9.8 * 2²
Sin Ф = 1/2. So Ф = 30°.
Range = u² Sin2Ф / g = 1600 * Sin 60° / 9.8 m
R = 141.4 m
Given at t = 2 sec, y = 20.4 m,
y = u SinФ t - 1/2 g t²
20.4 = 80 SinФ - 1/2 * 9.8 * 2²
Sin Ф = 1/2. So Ф = 30°.
Range = u² Sin2Ф / g = 1600 * Sin 60° / 9.8 m
R = 141.4 m
kvnmurty:
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