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An object is thrown with initial speed 5ms-1 with an angle of projection 30°calculate the maximum height reached and the horizontal angle |
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Answer» HEIGHT =0.3125 m range = 2.165 m Explanation: height = (U*sin(theta))^2/2g = 25/4 * 1/20 =0.3125 range = u^2*sin(2*theta)/g = 25*√3/2*1/10=2.165 |
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