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 :Here, `v_(0) = <a href="https://interviewquestions.tuteehub.com/tag/240-295938" style="font-weight:bold;" target="_blank" title="Click to know more about 240">240</a> ms^(-<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 = 5.10 km = 5100 m`, <br/> `g = 9.8 ms^(-2), <a href="https://interviewquestions.tuteehub.com/tag/alpha-858274" style="font-weight:bold;" target="_blank" title="Click to know more about ALPHA">ALPHA</a> = ? ""R = (v_(0)^(2) sin 2 alpha)/(g)` <br/> `sin 2 alpha = (Rg)/(v_(0)^(2)) ""rArr alpha = 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/> using `= T = (2v_(0)sin alpha)/(g)` <br/> when, `alpha = 30^(@) , T_(1) = (2 xx 240 xx 0.5)/(9.8) = 24.5 s` <br/> When, `alpha = 60^(@), T_(2) = (2 xx 240 xx 0.867)/(9.8) = 42.41s`</body></html>