Let's break down the problem step by step: --- ## **Step 1: List Given Data** - Angle of launch, \(\theta = 35^\circ\) - Height of ramp, \(h = 1.13\, \text{m}\) - Initial velocity, \(v_ = 2.48\, \text{m/s}\) - Acceleration due to gravity, \(g = 9.8\, \text{m/s}^2\) --- ## **Step 2: Find Initial Velocity Components** - **Horizontal component:** \(v_{x} = v_ \cos\theta\) - **Vertical component:** \(v_{y} = v_ \sin\theta\) \[ v_{x} = 2.48 \cos(35^\circ) \approx 2.03\, \text{m/s} \] \[ v_{y} = 2.48 \sin(35^\circ) \approx 1.42\, \text{m/s} \] --- ## **Step 3: Find Time of Flight** Use the vertical motion equation (taking downward as positive): \[ y = v_{y} t + \frac{1}{2}(-g)t^2 \] Since the marble lands on the ground, \(y = -1.13\, \text{m}\): \[ -1.13 = 1.42 t - \frac{1}{2}(9.8)t^2 \] \[ -1.13 = 1.42t - 4.9t^2 \] \[ 4.9t^2 - 1.42t - 1.13 = \] Solve this quadratic equation for \(t\): \[ t = \frac{-(-1.42) \pm \sqrt{(-1.42)^2 - 4 \cdot 4.9 \cdot (-1.13)}}{2 \cdot 4.9} \] \[ t = \frac{1.42 \pm \sqrt{2.0164 + 22.148}}{9.8} \] \[ t = \frac{1.42 \pm \sqrt{24.1644}}{9.8} \] \[ t = \frac{1.42 \pm 4.92}{9.8} \] Take the positive root: \[ t = \frac{1.42 + 4.92}{9.8} = \frac{6.34}{9.8} \approx .647\, \text{seconds} \] --- ## **Step 4: Find Horizontal Distance (Range)** \[ \text{Range} = v_{x} \cdot t \] \[ = 2.03\, \text{m/s} \cdot .647\, \text{s} \approx 1.31\, \text{m} \] --- ## **Final Answer** \[ \boxed{1.31\ \text{meters}} \] The marble traveled **1.31 meters** horizontally from the end of the ramp to where it landed.