The height `y` and distance `x` along the horizontal plane of a projectile on a certain planet are given by `x = 6t m` and `y = (8t - 5t^(2))m`. The velocity with which the projectile is projected is

Options

A. `8 ms^(-1)`
B. `9 ms^(-1)`
C. `10 ms^(-1)`
D. `(10//3) ms^(-1)`

Explanation

To find the velocity with which the projectile is projected, we will follow these steps:

### Step 1: Identify the equations of motion
The equations given for the projectile's motion are:
- Horizontal distance: \( x = 6t \) (in meters)
- Vertical height: \( y = 8t - 5t^2 \) (in meters)

### Step 2: Determine the x-component of velocity
The x-component of velocity (\( v_x \)) is the derivative of the horizontal distance with respect to time:
\[ v_x = \frac{dx}{dt} = \frac{d(6t)}{dt} = 6 \text{ m/s} \]

### Step 3: Determine the y-component of velocity
The y-component of velocity (\( v_y \)) is the derivative of the vertical height with respect to time:
\[ v_y = \frac{dy}{dt} = \frac{d(8t - 5t^2)}{dt} = 8 - 10t \text{ m/s} \]

### Step 4: Calculate the y-component of velocity at \( t = 0 \)
To find the initial velocity of the projectile, we substitute \( t = 0 \) into the equation for \( v_y \):
\[ v_y(t=0) = 8 - 10(0) = 8 \text{ m/s} \]

### Step 5: Calculate the resultant velocity
The resultant velocity (\( v \)) can be found using the Pythagorean theorem, since the x and y components are perpendicular to each other:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ m/s} \]

### Final Answer
The velocity with which the projectile is projected is \( 10 \text{ m/s} \).

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