A uniform solid spherical ball is rolling down a smooth inclined plane from a height h. The velocity attained by the ball when it reaches the bottom of the inclined plane is v. If ball when it reaches the bottom of the inclined plane is v. If the ball is now thrown vertically upwards with the same velocity `v`, the maximum height to which the ball will rise is

Options

A. `(5h)/(8)`
B. `(3h)/(5)`
C. `(5h)/(7)`
D. `(7h)/(9)`

Explanation

To find the maximum height to which a uniform solid spherical ball will rise when thrown vertically upwards with a velocity \( v \), we can follow these steps:

### Step 1: Understand the energy conversion
When the ball rolls down the inclined plane, it converts its potential energy at height \( h \) into kinetic energy at the bottom. The total kinetic energy of a rolling object is the sum of its translational and rotational kinetic energy.

### Step 2: Write the expression for potential energy
The potential energy (PE) at height \( h \) is given by:
\[ PE = mgh \]

### Step 3: Write the expression for kinetic energy
For a uniform solid sphere rolling without slipping, the total kinetic energy (KE) is given by:
\[ KE = KE_{translational} + KE_{rotational} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia of the sphere, and \( \omega \) is the angular velocity. For a solid sphere, \( I = \frac{2}{5} m r^2 \) and \( v = r \omega \).

### Step 4: Substitute and simplify the kinetic energy
Substituting \( I \) and \( \omega \):
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right \left(\frac{v}{r}\right)^2 \]
\[ = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2 \]

### Step 5: Set potential energy equal to kinetic energy
At the bottom of the inclined plane, the potential energy converts into kinetic energy:
\[ mgh = \frac{7}{10} mv^2 \]
Cancel \( m \) from both sides:
\[ gh = \frac{7}{10} v^2 \]

### Step 6: Solve for \( v^2 \)
Rearranging gives:
\[ v^2 = \frac{10}{7} gh \]

### Step 7: Use the velocity for vertical projection
When the ball is thrown vertically upwards with velocity \( v \), we can use the kinematic equation for maximum height:
\[ v^2 = u^2 + 2gh' \]
where \( u = v \) (initial velocity) and \( h' \) is the maximum height reached. Setting \( u^2 = v^2 \):
\[ \frac{10}{7} gh = 0 + 2gh' \]

### Step 8: Solve for \( h' \)
Rearranging gives:
\[ h' = \frac{10}{14} h = \frac{5}{7} h \]

### Final Answer
The maximum height to which the ball will rise is:
\[ h' = \frac{5}{7} h \]

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