Shouldn't he reach the top?

During the entire motion, mechanical energy is conserved. Assigning $y = 0$ to the center of the circle, we have $E_0 = mgR$ , so that $\tfrac12 m v^2 = mg(R - y).$ As long as the skier is on the circular track, the radial component of the net force is $F_{net,rad} = \frac{mv^2}R = \frac{2mg(R - y)}R.$ This radial force component is due to the normal force from the track, and the radial component of the gravitational force $mg\sin \theta = mgy/R$ . The skier will lose contact when the normal force changes from inward to outward; at that moment, the normal force is zero. Thus we obtain the equation $\frac{2mg(R - y)}R = \frac{mgy}R \\ 2(R-y) = y \\ y = \frac23 R.$ Once the skier is "launched", we have projectile motion. At the highest point, the speed is equal to the horizontal velocity component at the launch. This is equal to $v_x = v\sin \theta = \sqrt{2g(R - y)}\frac y R = \sqrt{\frac 23 Rg}\cdot \frac 23 = 30\cdot \frac 23 = \boxed{20}.$

Let $y$ be the height of the object from the ground. At first, $y=h$ . When the object is moving up the circular track, the magnitude of the centripetal acceleration is

$a_{c} =a_{g} +a_{N}$ ,

where $a_{g}$ is the radial component of the gravitational acceleration, and $a_{N}$ is the acceleration due to the normal force of the track acting on the object. By the law of conservation of energy,

$v= \sqrt{2g(h-y)}$ .

Moreover, the centripetal acceleration of an object undergoing circular motion is

$a_{c} =\frac{v^{2}}{r}$

When the object is at the upper half of the circular track, the radial component of gravitational acceleration is

$a_{g} =g(\frac{y-r}{r})$

Hence, the expression for the acceleration due to normal force is

$a_{N}=a_{c} - a_{g}=\frac{v^{2}}{r}-g(\frac{y-r}{r})=\frac{2g(h-y)}{r}-g(\frac{y-r}{r})$

When the object just loses contact with the inner surface of the track, $a_{N}=0$ . Thus,

$\frac{2g(h-y)}{r}-g(\frac{y-r}{r})=0$

$y=\frac{5h}{6}, v=\sqrt{2g(h-\frac{5h}{6})}=\sqrt{\frac{gh}{3}}\approx3$

From this point onward, the object will follow a parabolic trajectory. At the maximum height, its speed is just the horizontal component of the tangential velocity when it first left the surface of the track, $v_{x}$ . When $y=\frac{5h}{6}$ , $\frac{v_{x}}{v}=\frac{y-r}{r}=\frac{2}{3}$ . Therefore,

$v_{x}=\boxed{20}$

Let velocity be $v$ and angle be $\alpha$ (w.r.t horizontal) when when it leaves the track.we can find the expression for v from energy conservation and get $1/2mv^2 + mgh/2(1+sin \alpha) = mgh$ from here we can get an equation for $v$ .

Also when it leaves the track $N=0$ and force analysis yields $mg sin \alpha = mv^2/R = 2mv^2/h$ from here we get another equation for $v$ .

Solving these 2 equations we can get $sin\alpha$ and the value of $v$ .

Velocity at top is simply $vcos(\pi/2 - \alpha) = v sin \alpha$ $= 20 m/s$ (after putting $h=270$ and $g=10$ ).

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Bonus (proposed by me): Approximately what percentage of its original height does it reach after losing contact with the loop ?

Using Conservation of Energy, we get the velocity of the man at the bottom of the arc to be $u = \sqrt{2gH}$

where $H = 270 \text{m}$

$u = \sqrt{5400}$

When the skier is moving on the arc, we notice that

$N - mg \cos{\theta} = \frac{mv^2}{R}\ \ \ \ \ \ \ \$ $\color{#3D99F6}{(1)}$

where $N$ is the normal reaction by the wall on the skier, $\theta$ is the angle made by the radius joining the skier with the vertical and $v$ is the velocity of the skier at that instant.

Also, By Work Energy Theorem, the velocity $v$ is given by

$v = \sqrt{u^2 - 2gR(1-\cos{\theta})} \ \ \ \ \ \ \ \$ $\color{#3D99F6}{(2)}$

When the skier breaks off the surface, $N \rightarrow 0$ . Using this limiting case in $\color{#3D99F6}{(1)}$ and $\color{#3D99F6}{(2)}$ to find the value of $\theta$ just before the skier breaks off the surface, we get

$\theta = \cos^{-1}(\frac{-2}{3})$

Substituting the above in $\color{#3D99F6}{(2)}$ we get

$v = 30 \frac{m}{s}$

After this point the skier exhibits projectile motion. The angle velocity $v$ makes with horizontal is found to be $(\pi - \theta)$ . We know that at the highest point of a projectile, it possess only its horizontal component of velocity = $v \cos{(\pi - \theta)} = 30 \times \frac{2}{3} = \boxed{20 \ \tfrac{m}{s}})$

Ayyyy, this problem was really fun! Anyways, jumping right into the solution, the most important realization while solving this problem comes from realizing that the skier must have some amount of kinetic energy to supply centripetal acceleration when travelling around the loop. The skier will have initial energy:

$E_0 \ = \ mgH_0 \ = \ 2mgR$

Now, while the skier is in the bottom half of the semi-circle, the normal force can provide any amount of centripetal acceleration the system requires, so we're all good there. Once the ball reaches a vertical distance of $R$ inside the semi-circle, gravity will supply centripetal acceleration. This means that, in general, we must have:

$\frac{v^2}{R} \ \geq \ g \ \sin \ \theta \ \Rightarrow \ v \ \geq \ \sqrt{Rg\sin \ \theta}$

Where $\theta$ is the angle between the horizontal and the line passing through the centre of the semi-circle and the skier. This can be found by some pretty simple trigonometry. Now, we have the initial energy of the system. We need to find the angle $\theta$ such that the velocity is minimized but the centripetal acceleration condition is still satisfied (this is the point where the skier will lose contact with the circular loop). We have:

$2mgR \ = \ mgR \ + \ mgR \sin \ \theta \ + \ \frac{1}{2} mv^2 \ = \ mgR(1 \ + \ \sin \ \theta) \ + \ \frac{1}{2} \ Rmg \ \sin \ \theta \ \Rightarrow \ \sin \ \theta \ = \ \frac{2}{3}$

Notice that we have the gravitational potential of the skier inside the circular loop as $mgR(1 \ + \ \sin \ \theta)$ . This was again determined by some simple trigonometry. The skier's velocity when we leaves the loop is then:

$v \ = \ \sqrt{Rg\sin \ \theta} \ = \ \sqrt{\frac{2}{3} \ Rg}$

Now, finally, we know that when the ball leaves the loop, it will travel as a projectile. At it's maximum height, we know that $v_y \ = \ 0$ , so we have to calculate $v_x \ = \ v \ cos \ \phi$ , where $\phi$ is the launch angle of the skier. Well, we know $\theta$ when the skier leaves the surface of the loop, and using simple trigonometry, we can find $\ cos \ \phi \ = \ \sin \ \theta$ . Therefore, the skier's velocity at has maximum height is:

$v_{max} \ = \ sin \ \theta \ \sqrt{\frac{2}{3} \ Rg} \ = \ \sqrt{\frac{8}{27} \ (135) (10)} \ = \ \sqrt{400} \ = \ 20 \ \text{m}/\text{s}$

i got a quite stupid solution the skier slide down the slope with acceleration due to gravity (-10) so his vertical velocity should be 270 x -10 =-2700 and the angular velocity is w=v/r so w = -2700/135 =-20