Question

A cylindrical pipe of radius r is rolling towards a frog sitting on the horizontal ground. Center of the pipe is moving with velocity v. To save itself ,the frog jumps off and passes over the pipe touching it only at top. Denoting air time of frog by T, horizontal range of the jump by R and acceleration due to gravity by g. Find T and R.

Joshi sir comment

According to given condition 2r = uy2/2g,  T= 2uy/g =24gr/g=4rg At top point due to touch, vecocity of the frog be ux2v. R=[uxuy/g]+[ux2v)uy/g]=[uy/g](2ux2v) =[2uy/g](uxv) =T(uxv) Now minimum possible value of velocity of frog at top to cross the cylinder is  gr (By the concept of vertical circle) now solve carefully without error

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