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Prove that horizontal range of projectile is same, whenfired at an angle θ and 90-0 |
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Answer» Horizontal range is given by R=u^2sin2A/gsin(180-2A)=sin2Asin2(90-A)=sin2Asin2B=sin2A...where B=90-ANow,R=u^2sin2A/g and R=u^2sin2B/gSince sin2A=sin2Bhencesame for A and B=90-A |
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