(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.
![enter image source here]()
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
