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 provided are:
- Horizontal distance: \( x = 3t \) (in meters)
- Vertical height: \( y = 4t - 6t^2 \) (in meters)

### Step 2: Determine the horizontal and vertical components of velocity
From the equation for horizontal distance \( x = 3t \), we can find the horizontal component of velocity:
\[ v_x = \frac{dx}{dt} = \frac{d(3t)}{dt} = 3 \, \text{m/s} \]

For the vertical distance \( y = 4t - 6t^2 \), we can find the vertical component of velocity:
\[ v_y = \frac{dy}{dt} = \frac{d(4t - 6t^2)}{dt} = 4 - 12t \, \text{m/s} \]

### Step 3: Calculate the vertical velocity at the time of projection
At the moment of projection (when \( t = 0 \)):
\[ v_y = 4 - 12(0) = 4 \, \text{m/s} \]

### Step 4: Find the magnitude of the resultant velocity (speed of projection)
The speed of projection \( u \) can be calculated using the Pythagorean theorem, as the horizontal and vertical components are perpendicular to each other:
\[ u = \sqrt{v_x^2 + v_y^2} = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s} \]

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

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