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A cannon and a targetare 5.10Km apartand located at the same level. How soon will the shell launchedwiththe initial velocity 240m/s reach the targetin the absence of airdrag? |
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Answer» Solution :Here, `v_(0) = 240 MS^(-1), R = 5.10 km= 5100 m` `g=9.8 ms^(-2), alpha`=? `R=(v_(0)^(2) sin 2alpha)/g` `sin 2alpha =(RG)/v_(0)^(2) rArr alpha = 30^(@)` or `60^(@)` using `=T=(2v_(0)sin alpha)/g` When, `a = 30^(@), T_(1) =(2 xx 240 xx 0.5)/(9.8) = 24.5 s` When, `alpha =60^(@), T_(2) =(2 xx 240 xx 0.867)/9.8 = 42.41` s |
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