A projectile is given an initial velocity of `(hat(i)+2hat(j))` The Cartesian equation of its path is `(g = 10 ms^(-1))` ( Here , `hati` is the unit vector along horizontal and `hatj` is unit vector vertically upwards)

Options

A. `y = 2x - 5x^2`
B. `y = x - 5x^2`
C. `4y=2x-5x^2`
D. `y=2x-25x^2`

Answer

To find the Cartesian equation of the projectile's path given its initial velocity, we can follow these steps:

### Step 1: Identify Initial Velocities
The initial velocity of the projectile is given as:
- \( u_x = 1 \, \text{m/s} \) (horizontal component)
- \( u_y = 2 \, \text{m/s} \) (vertical component)

### Step 2: Write the Displacement Equations
For horizontal motion (x-direction):
- The horizontal displacement \( x \) is given by:
\[ x = u_x t \]
Since \( u_x = 1 \, \text{m/s} \):
\[ x = 1 \cdot t \implies t = x \]

For vertical motion (y-direction):
- The vertical displacement \( y \) is given by:
\[ y = u_y t - \frac{1}{2} g t^2 \]
Substituting \( u_y = 2 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):
\[ y = 2t - \frac{1}{2} \cdot 10 t^2 \]
Simplifying this gives:
\[ y = 2t - 5t^2 \]

### Step 3: Substitute \( t \) from the x-equation into the y-equation
Now we substitute \( t = x \) into the equation for \( y \):
\[ y = 2x - 5x^2 \]

### Step 4: Write the Cartesian Equation
Thus, the Cartesian equation of the projectile's path is:
\[ y = 2x - 5x^2 \]

### Summary
The Cartesian equation of the projectile's path is \( y = 2x - 5x^2 \).