A projectile is thrown at an angle of `30^(@)` with a velocity of 10m/s. the change in velocity during the time interval in which it reaches the highest point is

Options

A. 10 m/s
B. 5 m/s
C. `5sqrt3m//s`
D. `10sqrt3m//s`

Answer

To solve the problem of finding the change in velocity of a projectile at its highest point, we can follow these steps:

### Step 1: Resolve the Initial Velocity
The initial velocity \( v_i \) of the projectile is given as \( 10 \, \text{m/s} \) at an angle of \( 30^\circ \). We can resolve this velocity into its horizontal and vertical components.

- **Horizontal Component**:
\[ v_{ix} = v_i \cdot \cos(30^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \]

- **Vertical Component**:
\[ v_{iy} = v_i \cdot \sin(30^\circ) = 10 \cdot \frac{1}{2} = 5 \, \text{m/s} \]

### Step 2: Determine the Velocity at the Highest Point
At the highest point of the projectile's trajectory, the vertical component of the velocity becomes zero (i.e., \( v_{fy} = 0 \)). The horizontal component remains unchanged throughout the motion (i.e., \( v_{fx} = v_{ix} = 5\sqrt{3} \, \text{m/s} \)).

### Step 3: Calculate the Change in Velocity
The change in velocity \( \Delta v \) can be calculated using the formula:
\[ \Delta v = v_f - v_i \]
Where:
- \( v_f = v_{fx} \hat{i} + v_{fy} \hat{j} = 5\sqrt{3} \hat{i} + 0 \hat{j} \)
- \( v_i = v_{ix} \hat{i} + v_{iy} \hat{j} = 5\sqrt{3} \hat{i} + 5 \hat{j} \)

Now, substituting these into the equation:
\[ \Delta v = (5\sqrt{3} \hat{i} + 0 \hat{j}) - (5\sqrt{3} \hat{i} + 5 \hat{j}) \]
\[ \Delta v = (5\sqrt{3} - 5\sqrt{3}) \hat{i} + (0 - 5) \hat{j} \]
\[ \Delta v = 0 \hat{i} - 5 \hat{j} \]

### Step 4: Find the Magnitude of Change in Velocity
The magnitude of the change in velocity is given by:
\[ |\Delta v| = \sqrt{(0)^2 + (-5)^2} = \sqrt{25} = 5 \, \text{m/s} \]

Thus, the change in velocity during the time interval in which the projectile reaches the highest point is \( 5 \, \text{m/s} \).

### Final Answer
The change in velocity is \( 5 \, \text{m/s} \).

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