If the equation of motion of a projectile is `y=3x -1/8 x^(2)` the range and maximum height are respectively (y and x are in metres)

Options

A. 18 m and 24 m
B. 24 m and 18 m
C. 24 m and 6 m
D. 12 m and 9 m

Answer

To solve the problem of finding the range and maximum height of the projectile given by the equation \( y = 3x - \frac{1}{8} x^2 \), we will follow these steps:

### Step 1: Identify the equation of motion
The given equation of motion is:
\[ y = 3x - \frac{1}{8} x^2 \]

### Step 2: Find the range
To find the range, we need to determine the values of \( x \) when \( y = 0 \):
\[ 0 = 3x - \frac{1}{8} x^2 \]

### Step 3: Factor the equation
We can factor out \( x \) from the equation:
\[ x(3 - \frac{1}{8} x) = 0 \]
This gives us two solutions:
1. \( x = 0 \)
2. \( 3 - \frac{1}{8} x = 0 \)

### Step 4: Solve for \( x \)
From the second equation:
\[ 3 = \frac{1}{8} x \]
Multiplying both sides by 8:
\[ x = 24 \]
Thus, the range of the projectile is \( R = 24 \) meters.

### Step 5: Find the maximum height
The maximum height occurs at the vertex of the parabola described by the equation. The x-coordinate of the vertex can be found using the formula:
\[ x = -\frac{b}{2a} \]
where \( a = -\frac{1}{8} \) and \( b = 3 \):
\[ x = -\frac{3}{2 \times -\frac{1}{8}} = -\frac{3}{-\frac{1}{4}} = 12 \]

### Step 6: Substitute \( x \) back into the equation to find \( y \)
Now, we substitute \( x = 12 \) back into the original equation to find the maximum height \( y \):
\[ y = 3(12) - \frac{1}{8}(12^2) \]
Calculating this:
\[ y = 36 - \frac{1}{8}(144) = 36 - 18 = 18 \]
Thus, the maximum height \( H = 18 \) meters.

### Final Result
The range and maximum height of the projectile are:
\[ \text{Range} = 24 \text{ meters}, \quad \text{Maximum Height} = 18 \text{ meters} \]