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 velocity with which the projectile is projected at \( t = 0 \), we will follow these steps:

### Step 1: Identify the equations of motion
We are given two equations:
- The height \( y \) as a function of time \( t \):
\[ y = 8t - 5t^2 \]
- The horizontal distance \( x \) as a function of time \( t \):
\[ x = 6t \]

### Step 2: Differentiate the equations to find velocities
To find the velocities in both the x and y directions, we need to differentiate the equations with respect to time \( t \).

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

**For the vertical motion (y-direction):**
\[ v_y = \frac{dy}{dt} = \frac{d(8t - 5t^2)}{dt} = 8 - 10t \]

### Step 3: Evaluate the velocities at \( t = 0 \)
Now, we will evaluate both velocities at \( t = 0 \).

**For \( v_x \):**
\[ v_x = 6 \text{ m/s} \quad \text{(constant)} \]

**For \( v_y \):**
\[ v_y = 8 - 10(0) = 8 \text{ m/s} \]

### Step 4: Calculate the resultant velocity
The resultant velocity \( v \) can be found using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting the values we found:
\[ v = \sqrt{(6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ m/s} \]

### Conclusion
The velocity with which the projectile is projected at \( t = 0 \) is \( 10 \text{ m/s} \).

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