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 \( y = 12x - \frac{3}{4}x^2 \), we can follow these steps:

### Step 1: Set the equation of motion to zero
To find the range, we need to determine the points where the projectile returns to the same vertical level (ground level). This occurs when \( y = 0 \).

\[ 0 = 12x - \frac{3}{4}x^2 \]

### Step 2: Rearrange the equation
Rearranging the equation gives us:

\[ \frac{3}{4}x^2 - 12x = 0 \]

### Step 3: Factor the equation
We can factor out \( x \):

\[ x\left(\frac{3}{4}x - 12\right) = 0 \]

### Step 4: Solve for \( x \)
Setting each factor to zero gives us:

1. \( x = 0 \) (the launch point)
2. \( \frac{3}{4}x - 12 = 0 \)

Solving the second equation:

\[ \frac{3}{4}x = 12 \]

Multiplying both sides by \( \frac{4}{3} \):

\[ x = 12 \times \frac{4}{3} = 16 \]

### Step 5: Conclusion
The range of the projectile is the value of \( x \) when it returns to the ground level, which is:

\[ \text{Range} = 16 \text{ meters} \]