The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by `y = (8t-5t^(2))` meter and `x = 6t` meter , where t is in second. The velocity with which the projectile is projected is

Options

A. `8m//sec`
B. `6m//sec`
C. `10 m//sec`
D. Not obtainable from the data

Answer

To find the velocity with which the projectile is projected, we will analyze the given equations for height \( y \) and horizontal distance \( x \) as functions of time \( t \):

1. **Given Equations**:
- Vertical height: \( y = 8t - 5t^2 \)
- Horizontal distance: \( x = 6t \)

2. **Finding Vertical Velocity**:
- The vertical velocity \( v_y \) can be found by differentiating \( y \) with respect to \( t \):
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(8t - 5t^2) \]
- Differentiating gives:
\[ v_y = 8 - 10t \]

3. **Finding Initial Vertical Velocity**:
- The initial vertical velocity is found by evaluating \( v_y \) at \( t = 0 \):
\[ v_y(0) = 8 - 10(0) = 8 \, \text{m/s} \]

4. **Finding Horizontal Velocity**:
- The horizontal velocity \( v_x \) can be found by differentiating \( x \) with respect to \( t \):
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(6t) \]
- Differentiating gives:
\[ v_x = 6 \, \text{m/s} \]

5. **Finding the Resultant Velocity**:
- The resultant velocity \( v \) can be found using the Pythagorean theorem, since \( v_x \) and \( v_y \) are perpendicular:
\[ v = \sqrt{v_x^2 + v_y^2} \]
- Substituting the values:
\[ v = \sqrt{(6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{m/s} \]

6. **Conclusion**:
- The velocity with which the projectile is projected is \( 10 \, \text{m/s} \).