InterviewSolution
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InterviewSolution
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A stone is projected fromthepointof a groundin such adirection soas to hita birdon thetop of atelegraphpostof heighth and then attain the maximum height 2h above the ground . Ifat theinstant of projection,the bird were to fly away horizontally with a unifrom speed. Find the ratio between the horizontal velocitiesof thebird and stone, if thestone still hitsthe brid while ascending . |
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Answer» Solution :Letthe stoneis projected with a velcoityu at an ANGLE `theta`WITHTHE horizontal , we have ` (2h) = ((u sin theta)^(2))/(2g) oru sin theta = 2 sqrt(gh)` Supposet is thetimetaken by thestoneto reachthe HEIGHTH abovethe ground. Then `h = u sin theta t - (1)/(2) "gt"^(2) ( or) ("gt"^(2))/(2) - u sin theta+ h = 0` As we have`u sin theta = 2sqrt(gh) "" THEREFORE ("gt"^(2))/(2) - 2sqrt(gh) t+ h = 0` Solvingaboveequationfor t, we get `t = (2sqrt(gh) pm sqrt((2sqrt(gh))^(2) - 4 xx (g)/(2)h))/(2 xx (g)/(2)) = (2sqrt(gh) pm sqrt(2gh))/(g)` Let ` t_(1) = sqrt((2h)/(g))(sqrt(2)-1) and t_(2) = sqrt((2h)/(g))(sqrt(2)+1)` Where `t_(1)` and `t_(2)` correspond to P and Q in the figure. Supoosev is the horizontalvelcoityof the bird. Then`PQ = vt_(2)` For the motion of stone PQ =` u cos theta (t_(2) - t_(1)) "" therefore (t_(2) - t_(1))` whichgives `(v)/(u cos theta) = (t_(2) -t_(1))/(t_(2)) =(sqrt((2h)/(g)) (sqrt(2)+1) - (sqrt(2)-1))/(sqrt((2h)/(g))(sqrt(2)+1)) = (2)/(sqrt(2) +1)` 
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