Projectiles and conservation of energy (water slide problem)

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Homework Help Overview

The problem involves a water slide scenario where a person starts from rest at a certain height and lands in a pool, with specific distances and heights given. The context is rooted in the principles of projectile motion and conservation of energy, particularly focusing on the relationship between potential and kinetic energy in a frictionless environment.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of conservation of mechanical energy, questioning how to relate potential energy at the top of the slide to kinetic energy at the bottom. There is also consideration of kinematic equations to analyze the projectile motion after leaving the slide, particularly focusing on horizontal distance and time of fall.

Discussion Status

Some participants have provided insights into the energy conservation approach and the kinematic aspects of the problem. There is an ongoing exploration of how to connect the energy calculations with the projectile motion, while also addressing potential misunderstandings about the inclusion of work in the energy equation.

Contextual Notes

Participants note the assumption of a frictionless slide and the specific height and distance parameters provided in the problem. There is a focus on ensuring clarity regarding the definitions and roles of potential and kinetic energy in the context of this scenario.

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Homework Statement


The water slide shown in the figure ends at a height of 1.50 m above the pool. If the person starts from rest at point A and lands in the water at point B, which has a horizontal distance L = 2.58 m from the base of the slide, what is the height h of the water slide? (Assume the water slide is frictionless.)
08-23alt.gif

Homework Equations


PEi+KEi+W=PEf+KEf
PE=mgh
KE=(1/2)mv2
Vertical displacement after leaving slide: Dy=volt+(1/2)at2
Horizontal displacement after leaving slide Dx=vft

The Attempt at a Solution


PEi+KEi+W=PEf+KEf
PEi=KEf
mghi=(.5)mvf2
(9.81)h=(.5)vf2

I'm not really sure where to go from here. Any help would be greatly appreciated.
Thanks,
Will
 
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well v would be v=√(2gh).

It looks like the person leaves the slide horizontally. Use the kinematic equations, you are given the range L.
 
First let's consider the energy part of the problem by setting up an coordinate frame such that y=0 at the bottom of the slide. From this you can apply the conservation of mechanical energry theorem PE1+KE1=PE2+KE2 which will result in PE1=KE2. This is because we defined y=0 at the base of the slide and thus PE2=0 with respect to that frame. Now let's look at the projectile part of the problem. We know how far the person traveled in the 1.5m from the ground to the base of the slide, and we can theorem the time it takes an object to fall 1.5 under the influence of gravity. So we know distance in the x direction and the time it took to travel that distance, thus we know our x velocity Vx. Last we link the energy and projectile portions of the problem to obtain a solution. We can now do this because we know KE2 since we calculated that velocity Vx. If we know KE2 we know the potential energy PE1 at the top of the slide and thus the height.
 
Also note that work W should not appear in the conservation of mechanical energy equation. If you were referring to Wg the work due to gravity you doubled down on that term since Wg is already accounted for in terms of potential energy in the conservation of mechanical energy equation.
 

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