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. 18 m
B. 16 m
C. 12 m
D. 21.6 m

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 equation of motion to zero
The range of the projectile is determined by the points where the vertical position \( y \) is zero. Therefore, we set the equation of motion to zero:
\[ y = 12x - \frac{3}{4}x^2 = 0 \]

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

### Step 3: Solve for \( x \)
Now, we solve the second factor:
\[ 12 - \frac{3}{4}x = 0 \]
Rearranging gives:
\[ \frac{3}{4}x = 12 \]
Multiplying both sides by \( \frac{4}{3} \):
\[ x = 12 \times \frac{4}{3} = 16 \, \text{m} \]

### Step 4: Conclusion
Thus, the range \( R \) of the projectile is:
\[ R = 16 \, \text{m} \]