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(A) 11 sec(B) 23 sec(C) 38secFor a projectile the ratio of maximum height reached to the square of flight time is(A)5: 47.(B) 5: 2(D) 10 1en tho beight hl attained then angle of |
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Answer» H=U²Sin²θ/2gT=2Usinθ/gnow if we take the ratio ofmaximum height reached to the square of flight time:H/T²=U²Sin²θ/2g/4U²sin²θ/g²=g/8If we take g=10 m/s² thenH/T²=10/8=5/4 |
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