Vertical Circular Motion (M3)

Hiya, I'd appreciate if someone could help me with the last part of this question please

A toy 'speed-racing' track consists of a slope smoothly joined at A to a part of a vertical circular loop of track of radius a that rests upon the ground at B (see diagram). A small toy car is released from rest at point D on the slope. Assume the motion to be modelled by a point mass sliding around a smooth track. D is at height 2a above the ground.

a) Find the speed of the car when it reaches B.
b) Show that, if it is still in contact with the track at P, where angle COP = θ, then the speed v at P is given by $v^2 = 2ga(1 - \cos \theta)$
c) Draw a diagram showing the forces acting on the car at P. By using the component of 'F = ma' in the radial direction, find a formula for the reaction at R between the car and the track, and show that R vanishes when $\cos \theta = \frac{2}{3}$.
d) With the help of the equation in part (b), find the vertical component of the car's velocity at the point where R vanishes. Deduce that, after it has passed B, the maximum height of the car above the ground is $\frac{50}{27} a$

I'm fine with the the first three parts of the question but I can't manage to do part (d). I tried working out v from the formula and splitting it into components but that didn't seem to work. Would appreciate any help - thanks

So I got $v=\sqrt{\frac{2}{3}ga}$ then for the vertical component I think you just multiply it by cos theta as the velocity acts through the line of the tangent at P. For the last bit can't you just consider energy? The kinetic energy at b = potential energy at max height

EDIT: in fact I don't think that last bit will work, I'm not sure why though...

Working out v from the formula and then splitting into parts is the way to do it, but be careful about how you resolve the velocity into components - the velocity will be at a tangent to the circle, so you need to multiply by $\sin{\theta}$, not $\cos{\theta}$. Using trig identites to work out the value of $\sin{\theta}$ will give you the correct answer.

As for the last part, the important thing to note is that the car will leave the track when the reaction force is zero - i.e., when $\cos{\theta} = \frac{2}{3}$. Assuming the car is no longer affected by the track, how will it move?