# Question #65642

May 18, 2017

(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 $= m g h$ 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
$\frac{1}{2} m {U}^{2} = m g h '$
$\implies U = \sqrt{2 g h '}$ ........(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 $S \in \theta$ component of the velocity which becomes $= 0$ at that point.
Using the kinematic expression
${v}^{2} - {u}^{2} = 2 g h$
and inserting given values we get
${0}^{2} - {\left(U \sin \theta\right)}^{2} = 2 \left(- g\right) h$
Since gravity is acting against the velocity so it is deceleration and $-$ sign is used in front of g
$\implies {\left(U \sin \theta\right)}^{2} = \left(2 g h\right)$
$\implies h = \frac{{U}^{2} {\sin}^{2} \theta}{2 g}$
Using (1) we get
$\implies h = \frac{2 g h ' {\sin}^{2} \theta}{2 g}$
$\implies h = h ' {\sin}^{2} \theta$

From above expression we see that $h = h '$ only if $\theta = {90}^{\circ}$, else it is different.

(ii) As explained above at the highest point car will have only horizontal component, vertical being $= 0$.
Which is $= U \cos \theta$