Let's break down the problem step-by-step: --- ## **Problem Overview** You are jumping off a cliff at an angle of 37° from the ground at a speed of 1.5 m/s. Ignore air resistance. We are to find: 1. **How fast (in m/s) will you be going just before you hit the water?** 2. **How high is the cliff in meters?** 3. **Convert the velocity to miles/hour and the cliff height to feet.** --- ## **Step 1: How fast (in m/s) will you be going just before you hit the water?** This is a projectile motion problem. - Initial velocity (\(v_\)) = 1.5 m/s at 37° above the horizontal. - The vertical component of velocity at the start: \(v_{y} = v_ \sin\theta = 1.5 \times \sin(37^\circ)\) - The horizontal component: \(v_{x} = v_ \cos\theta = 1.5 \times \cos(37^\circ)\) \(\sin(37^\circ) \approx .6018\) \(\cos(37^\circ) \approx .7986\) So, \(v_{y} = 1.5 \times .6018 = .9027\) m/s \(v_{x} = 1.5 \times .7986 = 1.1979\) m/s After jumping, gravity acts downward. Let's denote **h** as the height of the cliff. The final vertical velocity just before hitting the water: \(v_y^2 = v_{y}^2 + 2g h\), where \(g = 9.8\) m/s². The horizontal velocity doesn't change (no air resistance): \(v_x = v_{x}\) The final speed is: \(v = \sqrt{v_x^2 + v_y^2}\) But we **need the height** before we can calculate \(v_y\). So, let's solve for the height next. --- ## **Step 2: How high is the cliff in meters?** We'll use the vertical motion equation: \(y = v_{y} t + \frac{1}{2} a t^2\) When you hit the water, \(y = -h\) (downward): \[ -h = v_{y} t - \frac{1}{2} g t^2 \] But we don't know \(t\) or \(h\) yet. Let's solve for \(t\) in terms of \(h\): \[ = v_{y} t - \frac{1}{2} g t^2 + h \] \[ \frac{1}{2} g t^2 - v_{y} t - h = \] This is a quadratic equation in \(t\): \[ t = \frac{v_{y} + \sqrt{v_{y}^2 + 2gh}}{g} \] But we don't know \(h\). **But the problem doesn't give us \(h\) or \(t\) directly!** But we can try a different approach. Since the initial speed is small, perhaps the cliff is tall compared to the jump. But the question seems to expect us to use energy conservation: ### **Using Energy Conservation** Potential energy lost = Kinetic energy gained (vertical direction): \[ mgh + \frac{1}{2} m v_^2 = \frac{1}{2} m v^2 \] So, the speed just before hitting the water: \[ v = \sqrt{v_^2 + 2gh} \] But since \(v_\) is at an angle, only the vertical component affects the vertical motion, but the total speed at the end will include both horizontal and vertical components. So, let's find the **time to fall**. The horizontal distance is not specified, so let's answer with the information given. --- ## **Step 3: Final Steps and Answers** Let's denote the cliff height as \(h\). ### **(a) How fast (in m/s) will you be going just before you hit the water?** \[ v = \sqrt{v_x^2 + v_y^2} \] - \(v_x = 1.1979\) m/s (constant) - \(v_{y} = .9027\) m/s (upward) The final vertical velocity: \[ v_y^2 = v_{y}^2 + 2gh \] \[ v_y = \sqrt{(.9027)^2 + 2 \times 9.8 \times h} \] So, the total speed just before hitting the water: \[ v = \sqrt{(1.1979)^2 + [(.9027)^2 + 2 \cdot 9.8 \cdot h]} \] --- ### **(b) How high is the cliff in meters?** If we assume you jump at 1.5 m/s at 37° and ignore air resistance, and you hit the water at time \(t\), we can solve for \(h\) if we know \(t\), but with only the given data, \(h\) remains a variable. --- ### **(c) Convert to mph and feet** - \(1 \text{ m/s} = 2.23694 \text{ mph}\) - \(1 \text{ m} = 3.281 \text{ ft}\) --- ## **Summary Table** | Quantity | Formula | Value (in terms of \(h\)) | |------------------|-----------------------------------------|----------------------------------| | Final speed \(v\)| \(\sqrt{(1.1979)^2 + (.9027)^2 + 2 \cdot 9.8 \cdot h}\) m/s | \(\sqrt{2.25 + 19.6h}\) m/s | | Cliff height \(h\)| — | Unknown, not enough data | | Final speed (mph)| \(v_{m/s} \times 2.23694\) | \(\sqrt{2.25 + 19.6h} \times 2.23694\) mph | | Cliff height (ft)| \(h_{m} \times 3.281\) | \(h \times 3.281\) ft | --- ## **Conclusion** **You need the height of the cliff (\(h\)) to calculate the exact numbers.** If you know the height, plug it into the formulas above to get your answers! ### **Should you jump or reason with the Griz fans?** - Jumping off a high cliff is dangerous—physics says your speed will be significant! - Reasoning with Griz fans might be safer (and less painful) than a risky jump! --- **If you have a value for the cliff height, let me know and I can substitute it directly into the formulas above!**