For finding the tangential and normal accelerations, we require an expression for the speed as a function of height y
vy=v0 and vx=ky
∴v=√v2x+v2y
=√v20+k2y2
Therefore, tangential acceleration,
at=dvdt=k2y√v20+k2y2dydt
=k2yv0√v20+k2y2
or at=k2y√1+k2y2v20
Now,
the total acceleration is, a=√(dvydt)2+(dvxdt)2
=dvxdt=kdydt=kv0
Normal acceleration, an=√a2−a2t
=kv0√1+(kyv0)2