Projectiles and conservation of energy (water slide problem)

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SUMMARY

The discussion focuses on solving a physics problem involving a water slide, where a person starts from rest at a height and lands in a pool. The key equations used include the conservation of mechanical energy, expressed as PEi + KEi = PEf + KEf, and the kinematic equations for projectile motion. The participant, Will, seeks assistance in determining the height of the slide, given that the final height is 1.50 m and the horizontal distance traveled is 2.58 m. The solution involves calculating the initial potential energy and relating it to the kinetic energy at the base of the slide to find the height.

PREREQUISITES
  • Understanding of conservation of mechanical energy principles
  • Familiarity with potential energy (PE = mgh) and kinetic energy (KE = (1/2)mv²)
  • Knowledge of kinematic equations for projectile motion
  • Ability to apply basic algebra to solve equations
NEXT STEPS
  • Study the derivation of projectile motion equations
  • Learn about energy conservation in frictionless systems
  • Explore the relationship between height and velocity in free fall
  • Practice solving similar physics problems involving energy and motion
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Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for practical examples of projectile motion and energy principles.

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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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