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Backyard waterslide - not bad. Not bad at all.

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You can actually use algebra for that, though that is super worst case scenario for the numbers.

Now for real numbers using actual physics, 3d fluid dynamics is barely a thing, if accuracy matters, even before taking into account the human body and impact.

Huh? The only component missing from Pulsar's work there is figuring out the coefficient of friction of the slide with water on it (and thus energy lost.)
 
Oh I did the math before my post, and his post...

Unless the huh is to the second part, in which case just google that.
 
You guys are over thinking this.

The real story is they had octuplets. They did trial by fire until they got it right. You notice they never show the far end of the pool? Bloodstains. This kid was the last one, fortunately they finally got it right. 😎
 
Why not just guesstimate it, toss a greased bag of sand down the slide and see where it lands? Too light? Tape two bags together.

Jeez. Why bust out the protractors when a trial or 2 is all that is needed.
 
Which is a problem that, quite literally, takes 2 minutes.

From memory.

PE = mgh.

KE = 1/2 * m * V^2.

Trig gets your X and Y component of velocity.

Use the Y component to determine flight time.

Use that flight time to calculate flight distance.

Bam. Done.

In fact, to continue on,

Since PE at the top = KE at the bottom, mgh = 1/2 M V ^ 2. So (and there's a lesson here) mass cancels out entirely. Leaving g * h = 1/2 V ^ 2.

So your exit velocity = sqrt ( 2 * g * h )....

Seriously. This stuff is cake.

For you and me? Sure. For a 12 year old kid? I don't think so.
 
I think someone in middle school would probably be able to do that. Erm... 12 is middle school right?

The US is seriously lacking in science and math and you think an average middle schooler could do this? I dunno man.

In any case . . . You guys are harsh. The kid is 12 and built a friggin awesome waterslide, and he even took into account launch angle when designing the thing. I was impressed. At his age I MIGHT have built the thing, but the launch angle would have been determined by trial by fire. Starting with the wimpy kid in the neighborhood.
 
Ok smarty pants. You do it and show your work. Take into account the effect of water on the coefficient of friction, a 115 lb person and 30 ft long pool. You know the drop angle and the height.

For what it's worth, I teach physics. The problem as described is nearly identical to a problem that someone posted to physics forums about 7 or 8 years ago. I kept a copy of that problem and have used it in high school physics ever since. It's the exact same problem with slightly different numbers. The coefficient of friction is likely less than the kinetic coefficient of friction of ice. (Since, as was explained for many years, ice was slippery because the pressure on it forced a tiny bit of the ice back into water, so you were sliding over a surface of water). Thus mu < 0.06. And, at a 75 degree angle, the normal force would only be 25.88% of his weight, making the frictional force nearly negligible down the slide. The frictional force would be approximately equal to 4 Newtons on the way down, using .03 as a mu_k, compared to a component of gravity parallel to the slide, equal to 96.59% of his weight, or about 495 Newtons.

I'm looking for the problem, but found these in the introductory physics homework section at physicsforums.com
https://www.physicsforums.com/threads/coefficient-of-friction-problem.221394/

Oh heck, just use Google and search the physicsforums for water slide problems with a ramp into a pool. There are too many of them for me to search. VERY common problem.

edit: of course, the friction, even with the small mu, would be much more significant on the curved section for the ramp at the bottom, as the normal force would be greater due to the centripetal force required to change his direction. Nonetheless, as pointed out above, it would contribute to him going a shorter distance, rather than a greater distance.
 
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