InterviewSolution
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InterviewSolution
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A cannon fires successively two shells with velocity v_(0) = 250 m//s, the first at the angle therefore= 60^(@) and the second at the angle q_(2) = 45^(@)to the horizontal, the azimuth being the same. Neglecting the air drag, find the time interval between firings leading to the collision of the shells. |
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Answer» SOLUTION :Let the first shell travel for time `t_(1)` and second fortime `t_(2)` , before they collide . Both the shells must travel equal horizontal and VERTICAL distances to collide. For horizontal motion `v_(0) cos theta_(1) .t_(1) = v_(0) cos theta_(2) t_(2)` `i.e., (t_(1))/(t_(2)) =(cos theta)/(cos theta_(2)) , (t_(1))/(t_(2)) = (cos 45^(@))/( cos 60^(@)) = (1//sqrt(2))/(1//2)t_(2)` `i.e.,(t_(1))= sqrt(2) t_(2)` For vertical motion , `v_(0) sin theta_(1). t_(1)- (1)/(2) "gt"_(1)^(2) = v_(0) sin theta_(2) . t_(2) = (1)/(2) "gt"_(2)^(2) (or) v_(0) sin 60^(@),t_(1) - (1)/(2)"gt"_(1)^(2) = v_(0) sin 45^(@) .t_(2) - (1)/(2) "gt"_(2)^(2)` (or) `v_(0) ( sin 60^(@) t_(1) - sin 45^(@)t_(2)) = (g)/(2) (t_(1)^(2) -t_(2)^(2))` Puttingvalues`250((sqrt(3))/(2)sqrt(2)t_(2)- (1)/(sqrt(2))t_(2)) = (9.8)/(2)(2t_(2)^(2) - t_(2)^(2))RARR 250((sqrt(6) - sqrt(2))/(2)) t_(2)= 4.9 t_(2)^(2)` `t_(2) = (250(2.45-1.41))/(2 xx 4.9) (or) t_(2) = (250xx 1.04)/(9.8)= 26.7 s rArr t_(1) = sqrt(2) t_(2) = 1.414 xx 26.7= 37.8s` Timeinterval `t_(1) - t_(2) = 37.8s - 26.7 s = 11.1 s` |
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