If `theta` is the angle between the velocity and acceleration of a projectile at a point of its path, its value when the projectile is at the highest point is :

Options

A. `0^(@)`
B. `180^(@)`
C. `90^(@)`
D. `45^(@)`

Answer

To solve the problem of finding the angle \( \theta \) between the velocity and acceleration of a projectile at its highest point, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Projectile Motion**:
A projectile is launched with an initial velocity \( u_0 \) at an angle \( \theta \) with respect to the horizontal. The motion can be broken down into horizontal and vertical components.

2. **Components of Velocity**:
- The horizontal component of the initial velocity is \( u_{x} = u_0 \cos \theta \).
- The vertical component of the initial velocity is \( u_{y} = u_0 \sin \theta \).

3. **Velocity at the Highest Point**:
At the highest point of the projectile's trajectory, the vertical component of the velocity becomes zero because the projectile momentarily stops moving upward before it starts descending. Therefore, at the highest point:
- \( v_{y} = 0 \)
- The horizontal component of the velocity remains unchanged: \( v_{x} = u_0 \cos \theta \).

4. **Acceleration of the Projectile**:
The only force acting on the projectile (ignoring air resistance) is gravity, which acts downwards with an acceleration \( g \). Thus, the acceleration vector \( \vec{a} \) is directed vertically downward.

5. **Direction of Velocity and Acceleration**:
- The velocity vector \( \vec{v} \) at the highest point is horizontal (along the x-axis).
- The acceleration vector \( \vec{a} \) is vertical (along the negative y-axis).

6. **Finding the Angle \( \theta \)**:
The angle \( \theta \) between the velocity vector and the acceleration vector can be determined. Since the velocity is horizontal and the acceleration is vertical, they are perpendicular to each other.

7. **Conclusion**:
Therefore, the angle \( \theta \) between the velocity and acceleration at the highest point of the projectile's path is:
\[ \theta = 90^\circ \]

### Final Answer:
The value of \( \theta \) when the projectile is at the highest point is \( 90^\circ \).

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