irodov

and about this question if we take tangetial and vertical components of accn we have

ma(t) = mgsin@cos#- kmgcos@

ma(x) = mgsin@ - kmgcos@

but k =tan@ convieniently!!!!!!!!!!!!!!![sry abt the spelling i suck at eng :P]

SO WE HAVE

a(t) = -a(x)!!!!!

or dv(t) = - dv(x)

integrate dv(t) frm v0 to v and dv(x) from 0 to vcos#

we get

v(t) = v0/(1+cos#) ....

hope its the answer!!!!!!!!!!!

can any 1 confirm ..........[my e-book is still gettin downloaded :P so i cant check answer!!]