Find the velocity of a projectile at the highest point, if it is projected with a speed ` 15 m s^(-1)` in the direction `45^(@)` above horizontla. (take g = ` 10 m s^(-2))`

Answer

To find the velocity of a projectile at its highest point when projected with a speed of 15 m/s at an angle of 45 degrees above the horizontal, we can follow these steps:

### Step 1: Determine the initial velocity components
The initial velocity \( u \) can be broken down into horizontal and vertical components using trigonometric functions:

- The horizontal component \( u_x \) is given by:
\[ u_x = u \cos(\theta) = 15 \cos(45^\circ) \]

- The vertical component \( u_y \) is given by:
\[ u_y = u \sin(\theta) = 15 \sin(45^\circ) \]

### Step 2: Calculate the components
Using the values for \( \cos(45^\circ) \) and \( \sin(45^\circ) \):
\[ \cos(45^\circ) = \sin(45^\circ) = \frac{1}{\sqrt{2}} \]

Substituting these values:
\[ u_x = 15 \cdot \frac{1}{\sqrt{2}} = \frac{15}{\sqrt{2}} \, \text{m/s} \]
\[ u_y = 15 \cdot \frac{1}{\sqrt{2}} = \frac{15}{\sqrt{2}} \, \text{m/s} \]

### Step 3: Analyze the motion at the highest point
At the highest point of the projectile's trajectory:
- The vertical component of the velocity \( v_y \) becomes 0 (as the projectile momentarily stops moving upward before descending).
- The horizontal component of the velocity \( v_x \) remains constant throughout the motion since there is no horizontal acceleration.

Thus, at the highest point:
\[ v_y = 0 \, \text{m/s} \]
\[ v_x = u_x = \frac{15}{\sqrt{2}} \, \text{m/s} \]

### Step 4: Calculate the total velocity at the highest point
The total velocity \( v \) at the highest point can be calculated using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Since \( v_y = 0 \):
\[ v = \sqrt{\left(\frac{15}{\sqrt{2}}\right)^2 + 0^2} = \frac{15}{\sqrt{2}} \, \text{m/s} \]

### Final Answer
The velocity of the projectile at the highest point is:
\[ v = \frac{15}{\sqrt{2}} \, \text{m/s} \approx 10.61 \, \text{m/s} \]