The height `y` nad 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) m` and `x = 6t m`, where `t` is in seconds. The velocity with which the projectile is projected at `t = 0` is.

Options

A. `8 m s^-1`
B. `6 m s^-1`
C. `10 m s^-1`
D. Not obtainable from the data.

Explanation

To find the initial velocity of the projectile at \( t = 0 \), we will analyze the given equations for height \( y \) and horizontal distance \( x \).

### Step 1: Determine the vertical velocity component
The vertical position \( y \) is given by the equation:
\[ y = 8t - 5t^2 \]
To find the vertical velocity \( v_y \), we differentiate \( y \) with respect to time \( t \):
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(8t - 5t^2) \]
Calculating the derivative:
\[ v_y = 8 - 10t \]
Now, we evaluate \( v_y \) at \( t = 0 \):
\[ v_y(0) = 8 - 10(0) = 8 \, \text{m/s} \]

### Step 2: Determine the horizontal velocity component
The horizontal position \( x \) is given by the equation:
\[ x = 6t \]
To find the horizontal velocity \( v_x \), we differentiate \( x \) with respect to time \( t \):
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(6t) \]
Calculating the derivative:
\[ v_x = 6 \]
Now, we evaluate \( v_x \) at \( t = 0 \):
\[ v_x(0) = 6 \, \text{m/s} \]

### Step 3: Calculate the resultant velocity
The initial velocity \( v \) of the projectile can be found using the Pythagorean theorem since the vertical and horizontal components are perpendicular:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting the values we found:
\[ v = \sqrt{(6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{m/s} \]

### Conclusion
The initial velocity with which the projectile is projected at \( t = 0 \) is:
\[ \boxed{10 \, \text{m/s}} \]