A projectile has

Options

A. minimum velocity at the point of projection and maximum at the maximum height
B. maximum at the point of projection and minimum at the maximum height
C. same velocity at any point in its path
D. zero velocity at the maximum height irrespective of the velocity of projection

Answer

To solve the problem regarding the velocities of a projectile at different points in its trajectory, we will analyze the situation using the principles of conservation of mechanical energy.

### Step-by-Step Solution:

1. **Understanding the Problem:**
We need to find the minimum and maximum velocities of a projectile at two specific points: at the point of projection (ground level) and at the maximum height.

2. **Initial Conditions:**
Let the initial velocity of the projectile at the point of projection be \( u \). At this point, the height \( h = 0 \).

3. **Applying Conservation of Mechanical Energy:**
The total mechanical energy at the point of projection (ground level) is entirely kinetic:
\[ E_i = K_i = \frac{1}{2} m u^2 \]
At the maximum height \( h = H \), the potential energy is maximum and kinetic energy is minimum:
\[ E_f = U_f + K_f = mgh + \frac{1}{2} m v^2 \]

4. **Setting Up the Energy Conservation Equation:**
According to the conservation of mechanical energy:
\[ E_i = E_f \]
Therefore, we have:
\[ \frac{1}{2} m u^2 = mgh + \frac{1}{2} m v^2 \]

5. **Simplifying the Equation:**
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} u^2 = gh + \frac{1}{2} v^2 \]
Rearranging gives:
\[ \frac{1}{2} u^2 - gh = \frac{1}{2} v^2 \]

6. **Solving for Velocity \( v \):**
Multiply the entire equation by 2:
\[ u^2 - 2gh = v^2 \]
Taking the square root gives us the velocity at height \( h \):
\[ v = \sqrt{u^2 - 2gh} \]

7. **Finding Maximum and Minimum Velocities:**
- **Maximum Velocity:** Occurs at the point of projection where \( h = 0 \):
\[ v_{\text{max}} = \sqrt{u^2 - 2g(0)} = \sqrt{u^2} = u \]
- **Minimum Velocity:** Occurs at maximum height \( h = H \):
\[ v_{\text{min}} = \sqrt{u^2 - 2gH} \]
Here, \( H \) is the maximum height reached by the projectile.

8. **Conclusion:**
- The maximum velocity \( v_{\text{max}} \) is at the point of projection (ground level) and equals \( u \).
- The minimum velocity \( v_{\text{min}} \) is at the maximum height \( H \).

### Final Answer:
- Maximum velocity at the point of projection and minimum velocity at maximum height.