A projectile is given an initial velocity of ` ( hat(i) + 2 hat (j) ) m//s`, where ` hat(i)` is along the ground and `hat (j)` is along the vertical . If ` g = 10 m//s^(2) `, the equation of its trajectory is :

Options

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

Explanation

To find the equation of the trajectory of a projectile given its initial velocity, we can follow these steps:

### Step 1: Identify the initial velocity components
The initial velocity of the projectile is given as \( \mathbf{u} = \hat{i} + 2\hat{j} \) m/s. We can identify the horizontal and vertical components of the initial velocity:
- \( u_x = 1 \) m/s (horizontal component)
- \( u_y = 2 \) m/s (vertical component)

### Step 2: Write the equations of motion
The equations of motion for a projectile can be expressed as:
- Horizontal motion: \( x = u_x t \)
- Vertical motion: \( y = u_y t - \frac{1}{2} g t^2 \)

Where \( g = 10 \) m/s² is the acceleration due to gravity.

### Step 3: Solve for time \( t \)
From the horizontal motion equation, we can express time \( t \) in terms of \( x \):
\[ t = \frac{x}{u_x} = \frac{x}{1} = x \text{ seconds} \]

### Step 4: Substitute \( t \) into the vertical motion equation
Now, substitute \( t = x \) into the vertical motion equation:
\[ y = u_y t - \frac{1}{2} g t^2 \]
Substituting the values:
\[ y = 2x - \frac{1}{2} \cdot 10 \cdot x^2 \]
\[ y = 2x - 5x^2 \]

### Step 5: Write the final equation of the trajectory
Thus, the equation of the trajectory of the projectile is:
\[ y = 2x - 5x^2 \]

### Summary of the solution
The equation of the trajectory for the given projectile is \( y = 2x - 5x^2 \).

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