IIT JEE , AIEEE Forum - Physics , Mechanics - QUE IN MECH... PLZ HELP ME OUT..RATES ASSURED..

varun.tinkle

4)
I AM ASSUMING THAT THE AT THE HIGHEST POINT OF THE TRAJECTORY IT CROSSES THE WALL THEN
LET THE HIEGHT OF THE WALL BE X
THE HORIZONTAL COMPONENT OF THE VELOCITY IS UCOS 45
THEREFORE THE TIME TAKEN TO CROSS A DIST A IS U COS 45/A
SINCE AT THIS POINT IT REACHES THE HIGHEST POINT OF ITS JOURNEY SO THE TIME TRAVELLED=1/2OF THE TOTAL TIME OF FLIGHT WHICH IS EQUAL TO USIN 45/G
USIN 45/G=U COS 45/A
THE MAX HIEGHT REACHED IS A
SO U^2SIN^245/2G=X
THE RANGE IS B
SO U^2SIN90/G=B
SOLVE THE EQUATIONS AND U WILL GET THE VALUE OF X

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