A ball strikes a smooth horizontal ground at an angle of `45^(@)` with the vertical. What cannot be the possible angle of its velocity with the vertical after the collision. (Assume e `le` 1).

Options

A. `45^@`
B. `30^@`
C. `53^@`
D. `60^@`

Answer

To solve the problem, we need to analyze the motion of a ball that strikes a smooth horizontal ground at an angle of 45 degrees with the vertical. We will determine the possible angles of its velocity with the vertical after the collision, given that the coefficient of restitution \( e \) is less than or equal to 1.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- The ball strikes the ground at an angle of \( 45^\circ \) with the vertical.
- This means the angle with the horizontal is \( 90^\circ - 45^\circ = 45^\circ \).

2. **Resolve Initial Velocity Components:**
- Let the initial velocity of the ball be \( V \).
- The vertical component of the velocity before the collision is:
\[ V_{\text{vertical}} = V \cos(45^\circ) = \frac{V}{\sqrt{2}} \]
- The horizontal component of the velocity before the collision is:
\[ V_{\text{horizontal}} = V \sin(45^\circ) = \frac{V}{\sqrt{2}} \]

3. **Understand the Collision:**
- The ground is smooth, meaning there is no friction. Therefore, the horizontal component of the velocity remains unchanged after the collision.
- The vertical component of the velocity after the collision will be affected by the coefficient of restitution \( e \):
\[ U_{\text{vertical}} = e \cdot V_{\text{vertical}} = e \cdot \frac{V}{\sqrt{2}} \]

4. **Determine the New Velocity Components:**
- After the collision, the horizontal component remains:
\[ U_{\text{horizontal}} = \frac{V}{\sqrt{2}} \]
- The new vertical component is:
\[ U_{\text{vertical}} = e \cdot \frac{V}{\sqrt{2}} \]

5. **Calculate the Angle After Collision:**
- The angle \( \alpha \) that the velocity makes with the vertical after the collision can be determined using the tangent function:
\[ \tan(\alpha) = \frac{U_{\text{horizontal}}}{U_{\text{vertical}}} = \frac{\frac{V}{\sqrt{2}}}{e \cdot \frac{V}{\sqrt{2}}} = \frac{1}{e} \]

6. **Analyze the Coefficient of Restitution:**
- Since \( e \leq 1 \), it follows that \( \frac{1}{e} \geq 1 \).
- Therefore, \( \tan(\alpha) \geq 1 \), which implies:
\[ \alpha \geq 45^\circ \]

7. **Evaluate the Given Options:**
- The possible angles given are \( 45^\circ, 30^\circ, 53^\circ, \) and \( 16^\circ \).
- Since \( \alpha \) must be greater than or equal to \( 45^\circ \), the angle \( 30^\circ \) cannot be achieved.

### Conclusion:
The angle that cannot be the possible angle of the ball's velocity with the vertical after the collision is **30 degrees**.