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

A cannon and a target are 5.10 km apart and located at the same level.

1.	A cannon and a target are 5.10 km apart and located at the same level. How soon will the shell launched with the initial velocity 240 m/s reach the target in the absence of air drag?
Answer» <html><body><p></p>Solution :Hence `v_(0)=240ms^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>),R=5.10ikm=5100m,,g=9.8ms^(-2), theta=?` <br/> From formula `R=(v_(0)^(2)sin 2 theta)/g` <br/> We have `sin 2 theta=(Rg)/(v_(0)^(2))`, Putting <a href="https://interviewquestions.tuteehub.com/tag/values-25920" style="font-weight:bold;" target="_blank" title="Click to know more about VALUES">VALUES</a> we get <br/> `sin 2 theta =(5100xx9.8)/(240xx240)=0.8677=(sqrt(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>))/2=sin60^(@)` or `sin 120^(@)impliestheta=30^(@)` or `<a href="https://interviewquestions.tuteehub.com/tag/60-328817" style="font-weight:bold;" target="_blank" title="Click to know more about 60">60</a>^(@)` <br/> From formula `T=(2v_(0)sin theta)/g` <br/> When `theta =30^(@),T_(1)=(2xx240xx0.5)/9.8=24.5`sec <br/> When `theta=60^(@),T_(2)=(2xxd240xx0.867)/9.8=42.14` sec</body></html>