A student skates up a ramp that makes an angle `30^(@)` with the horizontal . He/she starts ( as shown in the figure) at the bottom of the ramp with speed `v_(0)` and wants to turn around over a semicicular path xyz of radius R during which he/she reaches a maximum height h (at point ) form the ground as shown in the figure . Assume that the energy loss is negligble and the force required for this turn at the highest point is provided by his / her weight only. Then (g is the acceleration due to gravity).

Options

A. `v_(0)^(2) - 2 gh = (1)/(2) gR`
B. `v_(0)^(2) - 2gh = (sqrt(3))/(2)gR`
C. the centripetal force required at points x and z is zero
D. the centripetal force required is maximum at points x and z

Explanation

To solve the problem, we will analyze the situation step by step:

### Step 1: Understand the Energy Conservation
The student starts at the bottom of the ramp with speed \( v_0 \) and moves up to point \( y \) at height \( h \). We can use the principle of conservation of mechanical energy since energy loss is negligible.

**Energy at the bottom (initial):**
- Kinetic Energy (KE) = \( \frac{1}{2} mv_0^2 \)
- Potential Energy (PE) = 0 (taking the bottom as the reference point)

**Energy at point \( y \) (final):**
- Kinetic Energy (KE) = \( \frac{1}{2} mv^2 \)
- Potential Energy (PE) = \( mgh \)

According to the conservation of energy:
\[ \text{Total Energy at bottom} = \text{Total Energy at point } y \]
\[ \frac{1}{2} mv_0^2 = mgh + \frac{1}{2} mv^2 \]

### Step 2: Simplify the Equation
We can cancel \( m \) from both sides since it appears in all terms:
\[ \frac{1}{2} v_0^2 = gh + \frac{1}{2} v^2 \]

### Step 3: Rearranging the Equation
Rearranging gives us:
\[ \frac{1}{2} v_0^2 - gh = \frac{1}{2} v^2 \]
Multiplying through by 2:
\[ v_0^2 - 2gh = v^2 \]

### Step 4: Relate to Centripetal Force
At the highest point \( y \), the only force providing the centripetal force is the weight of the student. The centripetal force \( F_c \) required for circular motion is given by:
\[ F_c = \frac{mv^2}{R} \]
At the highest point, this centripetal force is provided by the weight:
\[ mg = \frac{mv^2}{R} \]
Cancelling \( m \) gives:
\[ g = \frac{v^2}{R} \]

### Step 5: Substitute \( v^2 \) from Energy Conservation
From our earlier equation, we have \( v^2 = v_0^2 - 2gh \). Substituting this into the centripetal force equation:
\[ g = \frac{v_0^2 - 2gh}{R} \]
Rearranging gives:
\[ gR = v_0^2 - 2gh \]
Thus:
\[ v_0^2 = gR + 2gh \]

### Step 6: Final Result
Now we can express the maximum height \( h \) in terms of \( v_0 \) and \( g \):
\[ v_0^2 = g(R + 2h) \]
This implies:
\[ h = \frac{v_0^2}{2g} - \frac{R}{2} \]