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A body is thrown with some velocity from the ground. Maximum height when it is thrown at `60^(@)` to horizontal is 90m. What is the height reached when it is thrown at `30^(@)` to the horizontal:-A. 90mB. 45mC. 30m/sD. 15m |
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Answer» Correct Answer - C `h=(u^(2)sin^(2)theta)/(2g)rArrh_(1)=90=((u sin 60^(@))^(2))/(2g)` `90=(u^(2))/(2g)xx[sqrt(3)/(2)]^(2)rArr(u^(2))/(2g)=120` `h_(2)=((u sin 30^(@))^(2))/(2g)=120xx(1)/(4)=30m` |
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