Solve Water Slide Problem: Initial Speed 0.54m/s, Height 3.2m

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The discussion centers on solving the water slide problem where a swimmer starts with an initial speed of 0.54 m/s from a height of 3.2 m. Participants recommend using the conservation of energy principle to calculate the final velocity at a height of 1.5 m, followed by kinematic equations to determine the horizontal distance traveled before splashing down. The relevant formulas include v = √(2gh) for velocity and h = (g*t²)/2 for vertical displacement. The discussion emphasizes neglecting air drag for accurate calculations.

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"If the height of the water slide is h = 3.2m and the person's initial speed at point A (at the top of slide) is 0.54m/s, at what location does the swimmer splash down in the pool?"

I need help with this guys. It may be relatively easy, but I'm not seeing it. I don't know what formula to apply here. I have an initial velocity and a height of the slide. Thanks in advance.
 
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Is that all the info given in the question? You can use conservation of energy to find the final velocity of the person, but from there I'm not sure what it is asking for.
 
It's asking for the distance from the end of the slide to the point where the swimmer splashes in the water.
 
How do I add an image?
 
In additional options, manage attachments. Well you know the Normal force will be 0 when the person is in the air about to splash.
 
This is a picture of the problem I'm having difficulty with. Thanks!
 

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Ok now with the picture is clearer, just find the speed at 1.5 m through conservation of energy and then work it out with kinematics (uniform acceleration neglecting air drag)
 
The "projectile" is launched from 1.5m above the pool,
with a horizontal velocity.
 
After getting the horizontal speed, you have to find the time that it takes to fall 1.5m (h=(g*t^2)/2) then you find the horizontal distance that the kid went by x=vt.
 
  • #10
Cyclovenom said:
Ok now with the picture is clearer, just find the speed at 1.5 m through conservation of energy and then work it out with kinematics (uniform acceleration neglecting air drag)

Would I use the formula v=square root of 2gh to figure out v2?
 

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