A ball is projected from the origin. The x- and y-coordinates of its displacement are given by `x = 3t and y = 4t - 5t^2`. Find the velocity of projection `("in" ms^(-1))`.

Explanation

To find the velocity of projection of the ball, we will follow these steps:

### Step 1: Identify the equations of motion
The displacement in the x and y directions are given by:
- \( x = 3t \)
- \( y = 4t - 5t^2 \)

### Step 2: Differentiate the displacement equations to find velocity components
To find the velocity components, we differentiate the displacement equations with respect to time \( t \).

1. For the x-component of velocity \( v_x \):
\[ v_x = \frac{dx}{dt} = \frac{d(3t)}{dt} = 3 \text{ m/s} \]

2. For the y-component of velocity \( v_y \):
\[ v_y = \frac{dy}{dt} = \frac{d(4t - 5t^2)}{dt} = 4 - 10t \text{ m/s} \]

### Step 3: Determine the velocity at the moment of projection
The ball is projected from the origin at \( t = 0 \). We will substitute \( t = 0 \) into the equation for \( v_y \):
\[ v_y = 4 - 10(0) = 4 \text{ m/s} \]

### Step 4: Calculate the magnitude of the velocity vector
The velocity vector at the moment of projection can be represented as:
\[ \vec{v} = (v_x, v_y) = (3, 4) \]
To find the magnitude of the velocity vector \( v \), we use the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ m/s} \]

### Conclusion
The velocity of projection is \( 5 \text{ m/s} \).

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