- || It’s been a great day of new, frictionless snow. Julie starts at the top of the 60° slope shown in FIGURE P10.44. At the bottom, a circular arc carries her through a 90° turn, and she then launches off a 3.0-m-high ramp. How far horizontally is her touchdown point from the end of the ramp?
The Correct Answer and Explanation is:
To solve this physics problem, we break it into three key stages:
- Motion down the slope
- Circular arc transition (which redirects motion)
- Projectile motion off the ramp
Step 1: Motion down the 60° slope
Assume Julie starts from rest, with no friction losses. Using energy conservation:
- Initial Energy (at the top) = Potential Energy Ei=mghE_i = mgh
- Final Energy (at the bottom of slope) = Kinetic Energy Ef=12mv2E_f = \frac{1}{2}mv^2
So, mgh=12mv2⇒v=2ghmgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh}
We need to determine the vertical height she descends. Assume she starts at a height hh. But we aren’t given a numerical value for hh, so we must calculate based on the angle and ramp geometry.
The problem indicates that she descends a 60° slope, goes through a 90° circular arc, then launches horizontally from a 3.0 m high ramp. That means her vertical drop (from the top to launch point) is 3.0 m, because after the arc she is 3.0 m above ground level.
So, total height fallen = 3.0 m
Therefore, v=2⋅9.8⋅3.0=58.8≈7.67 m/sv = \sqrt{2 \cdot 9.8 \cdot 3.0} = \sqrt{58.8} \approx 7.67 \, \text{m/s}
This is her horizontal velocity as she leaves the ramp.
Step 2: Projectile motion off the ramp
Now she is in free-fall with:
- Initial vertical velocity = 0 (since she launches horizontally)
- Initial horizontal velocity = 7.67 m/s
- Vertical height of fall = 3.0 m
Time to fall: y=12gt2⇒3.0=12(9.8)t2⇒t2=69.8⇒t≈0.78 sy = \frac{1}{2}gt^2 \Rightarrow 3.0 = \frac{1}{2}(9.8)t^2 \Rightarrow t^2 = \frac{6}{9.8} \Rightarrow t \approx 0.78 \, \text{s}
Horizontal distance: x=vxt=7.67⋅0.78≈6.0 mx = v_x t = 7.67 \cdot 0.78 \approx \boxed{6.0 \, \text{m}}
Final Answer:
Julie lands approximately 6.0 meters horizontally from the end of the ramp.
Explanation
This physics problem explores energy conservation and projectile motion principles. Julie starts from rest at the top of a frictionless 60° snow slope. Since friction is absent, mechanical energy is conserved — her potential energy at the top is completely converted into kinetic energy at the bottom.
The vertical height she descends is crucial. We are told she launches horizontally from a ramp 3.0 meters above the ground. Thus, the vertical drop from the start to launch point is 3.0 meters. Applying conservation of energy, her speed at the bottom (just before leaving the ramp) is derived from the equation v=2ghv = \sqrt{2gh}. Substituting g=9.8 m/s2g = 9.8 \, \text{m/s}^2 and h=3.0 mh = 3.0 \, \text{m}, we get v≈7.67 m/sv \approx 7.67 \, \text{m/s}.
Next, we analyze her flight after launching. Since she exits the ramp horizontally, her vertical velocity is zero at that point. The time it takes her to fall to the ground is determined solely by the vertical distance and gravity. Using the equation y=12gt2y = \frac{1}{2}gt^2, we solve for time and get approximately 0.78 seconds.
During this time, her horizontal speed remains constant because no forces act horizontally (neglecting air resistance). Multiplying her horizontal speed by time gives the horizontal range: x=vt=7.67⋅0.78≈6.0 mx = vt = 7.67 \cdot 0.78 \approx 6.0 \, \text{m}.
Thus, Julie lands 6.0 meters away from the end of the ramp. This result illustrates how energy conservation and projectile motion principles can accurately predict motion paths in frictionless conditions.
