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. `4y=2x-5x^(2)`
C. `4y=2x-25x^(2)`
D. `y=x-5x^(2)`

Answer

To find the equation of the trajectory of the projectile given the initial velocity and acceleration due to gravity, we can follow these steps:

### Step 1: Identify the initial velocity components
The initial velocity \( \vec{u} \) is given as:
\[ \vec{u} = \hat{i} + 2\hat{j} \, \text{m/s} \]
From this, we can identify:
- \( u_x = 1 \, \text{m/s} \) (horizontal component)
- \( u_y = 2 \, \text{m/s} \) (vertical component)

### Step 2: Write the equations of motion
The equations of motion for the projectile can be expressed as:
- For horizontal motion (x-direction):
\[ x = u_x \cdot t + \frac{1}{2} a_x t^2 \]
Since there is no horizontal acceleration (\( a_x = 0 \)):
\[ x = u_x \cdot t \implies x = 1 \cdot t \implies t = x \]

- For vertical motion (y-direction):
\[ y = u_y \cdot t + \frac{1}{2} a_y t^2 \]
Here, \( a_y = -g = -10 \, \text{m/s}^2 \):
\[ y = u_y \cdot t + \frac{1}{2} (-10) t^2 \]
Substituting \( u_y = 2 \):
\[ y = 2t - 5t^2 \]

### Step 3: Substitute \( t \) in the y-equation
From the horizontal motion, we found that \( t = x \). We can substitute this into the equation for \( y \):
\[ y = 2(x) - 5(x^2) \]
Thus, the equation of the trajectory becomes:
\[ y = 2x - 5x^2 \]

### Final Answer
The equation of the trajectory of the projectile is:
\[ y = 2x - 5x^2 \]

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