(i) Let #h'# be height of point #A# from where the toy car of mass #m# is released while at rest. As it reaches point #B#, at the bottom of curve, its potential energy #=mgh# got converted to its kinetic energy. (#B# is marked incorrectly in the figure).

If #U# is the velocity of the car at point #B#, ignoring frictional losses, and using law of Conservation of Energy we have

#1/2mU^2=mgh'#

#=>U=sqrt(2gh')# ........(1)

This velocity of the car at #B#, as it follows trajectory at angle of #theta#, can be resolved in its two horizontal and vertical components.

Maximum height #h# is reached because of #Sin theta# component of the velocity which becomes #=0# at that point.

Using the kinematic expression

#v^2-u^2=2gh#

and inserting given values we get

#0^2-(Usintheta)^2=2(-g)h#

Since gravity is acting against the velocity so it is deceleration and #-# sign is used in front of *g*

#=>(Usintheta)^2=(2gh)#

#=>h=(U^2sin^2theta)/(2g)#

Using (1) we get

#=>h=(2gh'sin^2theta)/(2g)#

#=>h=h'sin^2theta#

From above expression we see that #h=h'# only if #theta=90^@#, else it is different.

(ii) As explained above at the highest point car will have only horizontal component, vertical being #=0#.

Which is #=Ucostheta#