The horizontal and vertical displacements x and y of a projectile at a given time t are given by `x= 6 "t" and y= 8t -5t^2` meter.The range of the projectile in metre is:

Options

A. 9.6
B. 10.6
C. 19.2
D. 38.4

Explanation

To find the range of the projectile given the horizontal and vertical displacements, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Equations of Motion**:
The horizontal displacement \( x \) and vertical displacement \( y \) of the projectile are given by:
\[ x = 6t \]
\[ y = 8t - 5t^2 \]

2. **Determine the Time of Flight**:
The projectile will hit the ground when \( y = 0 \). We set the vertical displacement equation to zero:
\[ 0 = 8t - 5t^2 \]
Rearranging gives:
\[ 5t^2 - 8t = 0 \]
Factoring out \( t \):
\[ t(5t - 8) = 0 \]
This gives two solutions:
\[ t = 0 \quad \text{(initial point)} \]
\[ 5t - 8 = 0 \implies t = \frac{8}{5} \text{ seconds} \]

3. **Calculate the Range**:
The range \( R \) is the horizontal distance traveled when the projectile hits the ground. We substitute \( t = \frac{8}{5} \) into the horizontal displacement equation:
\[ x = 6t = 6 \left(\frac{8}{5}\right) \]
Simplifying this:
\[ x = \frac{48}{5} \text{ meters} \]

4. **Convert to Decimal**:
To express the range in decimal form:
\[ \frac{48}{5} = 9.6 \text{ meters} \]

5. **Final Answer**:
The range of the projectile is:
\[ \boxed{9.6 \text{ meters}} \]