Conservation of energy, ball kicked into a canyon

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Homework Help Overview

The problem involves a ball kicked off a 100.0 m canyon at an angle of 30.0 degrees with an initial velocity of 24.0 m/s, and participants are tasked with determining the maximum height reached above the canyon floor.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of conservation of energy principles and the need to resolve the initial velocity into horizontal and vertical components. There are questions about how to incorporate the angle into the calculations and what happens to the vertical component of velocity at the peak of the projectile's motion.

Discussion Status

The discussion is active, with participants exploring the relationship between the components of velocity and the conservation of energy. Some guidance has been provided regarding the conditions at the peak of the projectile's motion, and there is a focus on calculating the maximum height by considering both the initial height and the height gained during the projectile's ascent.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an emphasis on understanding the physics concepts rather than simply arriving at a numerical answer.

lking226
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Homework Statement


A ball is ikicked off a 100.0 m canyon at an angle of 30.0 degrees above the horizontal with a velocity of 24.0 m/s. What maximum height does it reach above the canyon floor?


Homework Equations


WNC = KE + PE


The Attempt at a Solution


WNC = 1/2mvf^2 - 1/2mvi^2 + mghf - mghi
1/2mvi^2 + mghi = 1/2mvf^2 + mghf

I'm not sure what to do with the 30.0 degree angle?
 
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lking226 said:
I'm not sure what to do with the 30.0 degree angle?

You can find the horizontal and vertical component of the initial speed of the ball
with the angle. When the ball is at its highest point, the vertical component will be 0.
 
lking226 said:

Homework Statement


A ball is ikicked off a 100.0 m canyon at an angle of 30.0 degrees above the horizontal with a velocity of 24.0 m/s. What maximum height does it reach above the canyon floor?


Homework Equations


WNC = KE + PE


The Attempt at a Solution


WNC = 1/2mvf^2 - 1/2mvi^2 + mghf - mghi
1/2mvi^2 + mghi = 1/2mvf^2 + mghf

I'm not sure what to do with the 30.0 degree angle?
You meant to say W_nc= delta KE + delta PE = 0, which is what you ended up with. When the ball reaches its max height, what is v_fy and what is v_fx?
 
okay so after i have vix = 20.785 and viy = 12, what do i do with those?
 
This is a projectile motion above a canyon. You've seem to have already found the vertical and horizontal components of the initial velocity so let's think about this some more.

In a projectile motion we're eventually going to reach a peak right?

Is there any specific property which notifies us that we've reached the peak of our projectile motion? (HINT: using the vertical/horizontal components of velcoity)
 
the velocity in the y direction will equal 0.
 
so i put that into the conservation of energy formula and solve for final height! thanks!
 
lking226 said:
so i put that into the conservation of energy formula and solve for final height! thanks!

I'm not entirely sure if that's correct. You told me above that the vertical component of its velocity will be 0 when it reaches its maximum height during its projectile motion. Calcuate this height and add it to the height of the ball initially above the canyon.

That will give you its maximum height.
 

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