The point from where a ball is projected is taken as the origin of the coordinate axes. The x and y components of its displacement are given by `x=6t and y=7t-5t^(2)`. What is the velocity of projection?

Options

A. `6ms^(-1)`
B. `8ms^(-1)`
C. `10ms^(-1)`
D. `14ms^(-1)`

Explanation

To find the velocity of projection of the ball, we need to analyze the given equations for the x and y components of displacement.

### Step 1: Identify the displacement equations
The displacement equations are given as:
- \( x = 6t \)
- \( y = 7t - 5t^2 \)

### Step 2: Calculate the velocity components
To find the velocity components, we need to differentiate the displacement equations with respect to time \( t \).

1. **For the x-component**:
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(6t) = 6 \text{ m/s} \]

2. **For the y-component**:
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(7t - 5t^2) = 7 - 10t \text{ m/s} \]

### Step 3: Find the velocity of projection
The velocity of projection is the initial velocity of the ball at \( t = 0 \). We need to evaluate \( v_y \) at \( t = 0 \).

1. **At \( t = 0 \)**:
\[ v_y(0) = 7 - 10(0) = 7 \text{ m/s} \]

2. **Thus, the initial velocity vector \( \vec{v} \)** can be expressed as:
\[ \vec{v} = (v_x, v_y) = (6, 7) \text{ m/s} \]

### Step 4: Calculate the magnitude of the velocity
To find the magnitude of the velocity of projection, we use the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{6^2 + 7^2} = \sqrt{36 + 49} = \sqrt{85} \text{ m/s} \]

### Conclusion
The velocity of projection is:
\[ v = \sqrt{85} \text{ m/s} \approx 9.22 \text{ m/s} \]

---