Conserve Energy? - 2

Let $H = 2R$ .

So, velocity of particle at the lowest point of the semicircle will be ${V}_{o} = \sqrt{4gR}$ . As we all know that this velocity is insufficient for the particle to go to the topmost point of the semicircle (The particle requires ${V}_{o} = \sqrt{5gR}$ to go to top). Hence it will leave the semicircle at some point and start projectile motion. Let this point make an angle $\theta$ with vertical and velocity of particle at this point be $V$ .

Since at this point particle just leave the semicircle so normal reaction given by the semicircle on the particle would be zero

$\implies$ $mgcos\theta = \frac{mV^{2}}{R}$

$\implies$ $gRcos\theta = V^{2}$ …………………….. (i)

By conservation of energy,

$\frac{1}{2}m{V_{o}}^{2} = mgR(1+cos\theta) + \frac{1}{2}mV^{2}$

$\implies$ $4gR = 2gR(1+cos\theta) + V^{2}$

$\implies$ $2gR = 3V^{2}$

or $\boxed{V = \sqrt{\frac{2gR}{3}}}$

and $\boxed{cos\theta = \frac{2}{3}}$

And velocity of particle at highest point of its trajectory would be $V_{max} = Vcos\theta = \frac{2}{3}\sqrt{\frac{2gR}{3}} = \frac{2}{3}\sqrt{\frac{gH}{3}}$ .

Now let $x$ be the maximum height reached by the particle.

Using Conservation of Enengy,

$mgh = mgx + \frac{1}{2}m{V_{max}}^{2}$

$\implies$ $x = \frac{25H}{27}$

Hence $\boxed{x = 5 m}$