The equation of motion of a projectile is `y = 12 x - (3)/(4) x^2`. The horizontal component of velocity is `3 ms^-1`. What is the range of the projectile ?

Options

A. 12m
B. 16m
C. 20m
D. 24m

Explanation

To find the range of the projectile given the equation of motion \( y = 12x - \frac{3}{4}x^2 \) and the horizontal component of velocity \( v_x = 3 \, \text{m/s} \), we can follow these steps:

### Step 1: Set the vertical position \( y \) to zero
The range of the projectile is the horizontal distance traveled when it returns to the ground level (where \( y = 0 \)). Therefore, we set the equation of motion equal to zero:
\[ 0 = 12x - \frac{3}{4}x^2 \]

### Step 2: Rearrange the equation
Rearranging the equation gives:
\[ \frac{3}{4}x^2 = 12x \]

### Step 3: Factor out \( x \)
We can factor out \( x \) from the equation:
\[ x\left(\frac{3}{4}x - 12\right) = 0 \]
This gives us two solutions:
1. \( x = 0 \) (the starting point)
2. \( \frac{3}{4}x - 12 = 0 \)

### Step 4: Solve for \( x \)
Now, we solve the second equation for \( x \):
\[ \frac{3}{4}x = 12 \]
Multiplying both sides by \( \frac{4}{3} \):
\[ x = 12 \cdot \frac{4}{3} = 16 \, \text{m} \]

### Step 5: Conclusion
Thus, the range of the projectile is:
\[ \text{Range} = 16 \, \text{meters} \]