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A cannon and a target are 5.10Km apart ...

A cannon and a target are 5.10Km apart and located at the same level. How soon will the shell launched with the initial velocity 240m/s reach the target in the absence of air drag?

Text Solution

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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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