The horizontal and vertical distances travelled by a particle in time t are given by x=6t and `y=8t-5t^(2)`. If g=`10m//sec^(2)`, then the initial velocity of the particle is

Options

A. 8m/sec
B. 10m/sec
C. 5m/sec
D. zero

Answer

To find the initial velocity of the particle given the horizontal and vertical distance equations, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the equations of motion**:
- The horizontal distance \( x \) is given by:
\[ x = 6t \]
- The vertical distance \( y \) is given by:
\[ y = 8t - 5t^2 \]

2. **Calculate the horizontal velocity**:
- The horizontal velocity \( v_x \) can be found by differentiating \( x \) with respect to time \( t \):
\[ v_x = \frac{dx}{dt} = \frac{d(6t)}{dt} = 6 \, \text{m/s} \]
- This value is constant and does not depend on time.

3. **Calculate the vertical velocity**:
- The vertical velocity \( v_y \) can be found by differentiating \( y \) with respect to time \( t \):
\[ v_y = \frac{dy}{dt} = \frac{d(8t - 5t^2)}{dt} = 8 - 10t \]
- At \( t = 0 \):
\[ v_y = 8 - 10(0) = 8 \, \text{m/s} \]

4. **Combine the horizontal and vertical velocities**:
- The initial velocity \( v \) of the particle is the vector sum of \( v_x \) and \( v_y \):
\[ v = \sqrt{v_x^2 + v_y^2} \]
- Substituting the values:
\[ v = \sqrt{(6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{m/s} \]

5. **Conclusion**:
- The initial velocity of the particle is:
\[ v = 10 \, \text{m/s} \]