The vertical height y and horizontal distance x of a projectile on a certain planet are given by `x= (3t) m, y= (4t-6t^(2))` m where t is in seconds, Find the speed of projection (in m/s).

Answer

To find the speed of projection of the projectile given the equations for horizontal distance \( x \) and vertical height \( y \), we can follow these steps:

### Step 1: Identify the equations
The equations given are:
- \( x = 3t \) (horizontal distance)
- \( y = 4t - 6t^2 \) (vertical height)

### Step 2: Differentiate \( x \) with respect to \( t \)
To find the horizontal component of the velocity (\( v_x \)), we differentiate \( x \) with respect to time \( t \):
\[ v_x = \frac{dx}{dt} = \frac{d(3t)}{dt} = 3 \, \text{m/s} \]

### Step 3: Differentiate \( y \) with respect to \( t \)
Next, we differentiate \( y \) with respect to time \( t \) to find the vertical component of the velocity (\( v_y \)):
\[ v_y = \frac{dy}{dt} = \frac{d(4t - 6t^2)}{dt} = 4 - 12t \, \text{m/s} \]

### Step 4: Evaluate \( v_y \) at \( t = 0 \)
To find the speed of projection, we evaluate \( v_y \) at \( t = 0 \):
\[ v_y(0) = 4 - 12(0) = 4 \, \text{m/s} \]

### Step 5: Calculate the resultant speed
The speed of projection is the magnitude of the velocity vector, which can be calculated using the Pythagorean theorem:
\[ \text{Speed} = \sqrt{v_x^2 + v_y^2} = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s} \]

### Final Answer
The speed of projection is \( 5 \, \text{m/s} \).

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