Calculating the Distance a Rock Travels Up a Sloped Hill

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To determine how far a rock travels up a frictionless hill with a slope of 23 degrees, the kinetic energy (KE) at the base must equal the potential energy (PE) at the highest point. The initial speed of the rock is 16 m/s, and the relevant equations are KE = 0.5mv² and PE = mgh, where h is the vertical height. The angle of the slope must be factored into the calculations to find the distance along the slope. The correct approach involves calculating the vertical height and then using trigonometry to find the distance traveled up the slope. Understanding the relationship between kinetic and potential energy is crucial for solving this problem.
eglaud

Homework Statement


A rock approaches the foot of a hill with a speed of 16 m/s. The hill has a slope of 23 degrees, how far up the hill does the rock travel?

Homework Equations


KE = PE

The Attempt at a Solution


I tried mv2/2 = mgy but I know this isn't right since I don't use theta at all. What do I do to change this?
 
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Is the equation KE = PE - W? This is what I did before, but I am unsure why except that I can plug in the angle, but I forgot to include this HILL IS FRICTIONLESS so I wouldn't have thought to use this.
 
eglaud said:

Homework Statement


A rock approaches the foot of a hill with a speed of 16 m/s. The hill has a slope of 23 degrees, how far up the hill does the rock travel?

Homework Equations


KE = PE

The Attempt at a Solution


I tried mv2/2 = mgy but I know this isn't right since I don't use theta at all. What do I do to change this?

Have you drawn a picture?
 
RedDelicious said:
Have you drawn a picture?

Yes, it looks like a rock moving towards a triangle
 
eglaud said:
how far up the hill does the rock travel?

eglaud said:
mv2/2 = mgy
Presumably your y is the vertical ascent, but I don't think that is quite what the question is asking for.
 
Hello everybody. Sorry for my english. I am not native. I think you must calculate de kinetic Energy at the bottom of the hill. Then, you must calculate the position (dependent with the angle) at the velocity is 0 m/s.
 
AgusCF said:
Hello everybody. Sorry for my english. I am not native. I think you must calculate de kinetic Energy at the bottom of the hill. Then, you must calculate the position (dependent with the angle) at the velocity is 0 m/s.
I think eglaud understands that but has not interpreted the question correctly. See my post #6.
 
haruspex said:
I think eglaud understands that but has not interpreted the question correctly. See my post #6.
Ouch! Yeap you are right! And I think that the problem wants to calculate the distance that the rock goes up on the lateral of the hill.
 

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