A ball of mass m is projected from the ground with an initial velocity u making an angle of `theta` with the vertical. What is the change in velocity between the point of projection and the highest point ?

Options

A. `u cos theta` downward
B. `u cos theta` upward
C. `u sin theta` upward
D. `u sin theta` downward

Explanation

To solve the problem, we need to find the change in velocity of a ball projected from the ground at an angle θ with the vertical, from the point of projection to the highest point of its trajectory.

### Step-by-step Solution:

1. **Identify the Components of Initial Velocity**:
- The initial velocity \( u \) can be broken down into two components:
- Vertical component: \( u_y = u \cos \theta \) (along the y-axis)
- Horizontal component: \( u_x = u \sin \theta \) (along the x-axis)

2. **Determine the Velocity at the Highest Point**:
- At the highest point of the projectile's path, the vertical component of the velocity becomes zero because the ball momentarily stops moving upward before it starts to descend.
- Therefore, the velocity at the highest point \( V_2 \) can be expressed as:
- \( V_2 = u_x \hat{i} + 0 \hat{j} = u \sin \theta \hat{i} \)

3. **Express the Initial Velocity**:
- The initial velocity \( V_1 \) at the point of projection can be expressed as:
- \( V_1 = u_x \hat{i} + u_y \hat{j} = u \sin \theta \hat{i} + u \cos \theta \hat{j} \)

4. **Calculate the Change in Velocity**:
- The change in velocity \( \Delta V \) is given by:
\[ \Delta V = V_2 - V_1 \]
- Substituting the expressions for \( V_2 \) and \( V_1 \):
\[ \Delta V = (u \sin \theta \hat{i}) - (u \sin \theta \hat{i} + u \cos \theta \hat{j}) \]
- Simplifying this gives:
\[ \Delta V = u \sin \theta \hat{i} - u \sin \theta \hat{i} - u \cos \theta \hat{j} = -u \cos \theta \hat{j} \]

5. **Interpret the Result**:
- The negative sign indicates that the change in velocity is directed downward. Thus, the change in velocity from the point of projection to the highest point is:
\[ \Delta V = -u \cos \theta \hat{j} \]

### Conclusion:
The change in velocity between the point of projection and the highest point is \( -u \cos \theta \hat{j} \), which means it is directed downward with a magnitude of \( u \cos \theta \).