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 can analyze the motion using the principles of physics, particularly the conservation of energy.

### Step-by-Step Solution:

1. **Understanding Projectile Motion**:
A projectile is an object that is thrown into the air with an initial velocity. It follows a curved path under the influence of gravity.

2. **Identifying Key Points**:
The key points in the projectile's motion are:
- The point of projection (initial point)
- The maximum height (the highest point in the trajectory)

3. **Applying Conservation of Energy**:
The total mechanical energy of the projectile is conserved if we ignore air resistance. This means that the sum of kinetic energy (KE) and potential energy (PE) remains constant throughout the motion.

- At the point of projection (initial point):
- Kinetic Energy (KE_initial) = \( \frac{1}{2} m u^2 \) (where \( u \) is the initial velocity)
- Potential Energy (PE_initial) = 0 (since we can take the height at this point as zero)

- At the maximum height:
- Kinetic Energy (KE_max_height) = \( \frac{1}{2} m v^2 \) (where \( v \) is the velocity at maximum height)
- Potential Energy (PE_max_height) = \( mgh \) (where \( h \) is the maximum height)

4. **Setting Up the Equation**:
According to the conservation of energy:
\[ KE_{initial} + PE_{initial} = KE_{max\ height} + PE_{max\ height} \]
This simplifies to:
\[ \frac{1}{2} m u^2 = \frac{1}{2} m v^2 + mgh \]

5. **Rearranging the Equation**:
Dividing through by \( m \) and rearranging gives:
\[ \frac{1}{2} u^2 = \frac{1}{2} v^2 + gh \]
Rearranging further:
\[ v^2 = u^2 - 2gh \]

6. **Analyzing the Results**:
- When \( h = 0 \) (at the point of projection), \( v^2 = u^2 \), hence \( v = u \). This indicates that the velocity is maximum at the point of projection.
- When \( h \) is at its maximum height, the velocity \( v \) will be minimum, and at the maximum height, the vertical component of the velocity becomes zero.

7. **Conclusion**:
Therefore, the correct statement is:
- Maximum velocity at the point of projection and minimum at the maximum height.

### Final Answer:
**Option 2**: Maximum at the point of projection and minimum at the maximum height.

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