Let's break down the problem: ### **Given Data** - Initial speed (\(v_\)): **12.4 m/s** - Launch angle (\(\theta\)): **40.°** - Initial height (\(y_\)): **1.80 m** - Acceleration due to gravity (\(g\)): **9.81 m/s²** - We need to find the **horizontal distance** (range, \(x\)). --- ## **Step 1: Find Initial Velocity Components** - \(v_{x} = v_ \cos\theta\) - \(v_{y} = v_ \sin\theta\) \[ v_{x} = 12.4 \cos(40.^\circ) = 12.4 \times .766 = 9.498 \text{ m/s} \] \[ v_{y} = 12.4 \sin(40.^\circ) = 12.4 \times .6428 = 7.969 \text{ m/s} \] --- ## **Step 2: Find Time of Flight** The vertical motion equation: \[ y = y_ + v_{y} t - \frac{1}{2}gt^2 \] At ground level, \(y = \): \[ = 1.80 + 7.969t - 4.905t^2 \] \[ 4.905t^2 - 7.969t - 1.80 = \] Use the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \(a = 4.905\), \(b = -7.969\), \(c = -1.80\): \[ t = \frac{7.969 \pm \sqrt{(-7.969)^2 - 4 \times 4.905 \times (-1.80)}}{2 \times 4.905} \] \[ t = \frac{7.969 \pm \sqrt{63.5 + 35.3}}{9.81} \] \[ t = \frac{7.969 \pm \sqrt{98.8}}{9.81} \] \[ t = \frac{7.969 \pm 9.94}{9.81} \] Only the positive root makes sense: \[ t = \frac{7.969 + 9.94}{9.81} = \frac{17.909}{9.81} \approx 1.83 \text{ s} \] --- ## **Step 3: Find the Range** \[ x = v_{x} \cdot t = 9.498 \times 1.83 \approx 17.4 \text{ m} \] --- ## **Final Answer** \[ \boxed{17.4 \ \text{m}} \] --- **The shot travels approximately 17.4 meters.**