I'm writing a children's book about hypothetical questions. One question is: how fast would you get to the bottom of the tallest possible slide in the world. This slide would be imaginary. I've made it 5 kilometres high, as few people survive for long above that height. I've assumed a 45 degree angle to the Earth. Also we can leave out any friction issues if it's easier (although we can leave them in if it makes for a more interesting answer). It's been a while since I studied physics so I'm having trouble working out a definitive answer.

  • 1
    $\begingroup$ I would argue Friction is the most important force here, as it starts to become increasingly important at higher speeds. If you ignore friction, then nothing is stopping you going arbitrarily fast! $\endgroup$ – Thomas Russell Apr 21 '17 at 12:58
  • 1
    $\begingroup$ Without friction, the speed in your specific example would be on the order of the speed of sound. $\endgroup$ – Chet Miller Apr 21 '17 at 13:28
  • 1
    $\begingroup$ all I know is this is the most dangerous bobsled course in Olympic history. And that says something $\endgroup$ – Jim Apr 21 '17 at 13:31

I'm not going to do the math for you; that would be cheating. However, I'm happy to help lead you to the answer.

For the acceleration due to gravity, you can assume it's constant at $g=9.8m/s^2$. I did the math on that one and the actual acceleration due to gravity changes very little and is right around that number the whole time, so it's a good approximation.

Now, $5km$ high at $45^{\circ}$ means it covers $5km$ along the ground. This is good because it means the effect of the curvature of Earth can be ignored. It also means that the resultant acceleration on your rider is $g\cos45=\frac{g}{\sqrt2}$

Now, you could go from here and factor in wind resistance to find the more or less accurate and precise velocity of your rider at the bottom, but I suggest we cheat. What I suggest you do is compute the velocity ignoring wind resistance, then just look up the approximate terminal velocity of a person. Figure out which of these two numbers is the smallest and then the answer to your question will be "Most definitely less than [lowest number]".

But how do you figure out the velocity ignoring air resistance? The easiest equation is: $$v^2=2ad$$

where $a$ is the acceleration along the slope of your slide ($\frac{g}{\sqrt2}$) and $d$ is the length of the slope of your slide (conveniently, this is $\sqrt2\cdot5km$). Solve for $v$.

Now that you've solved for $v$ and searched google for "approximate terminal velocity of person", you have two numbers. But why did I say the answer is most definitely less than the minimum of these? Because, if $v$ is less than terminal velocity, you know the rider can't even reach that without air resistance. However, you know air resistance WILL be a factor and will slow the rider down some. We haven't done the math, so we don't know exactly how much, but I promise you it will be significant. If, on the much more likely hand, terminal velocity is less than $v$, we know that, even if the slide were at a $90^{\circ}$ incline (that's technically more of a drop than a slide), the rider would not accelerate faster than terminal velocity. However, we also know that the acceleration this rider feels is less than $g$, which means their terminal velocity is also less. I'm not much of a fluid mechanist, but without doing a quick and easy search of google, I'm pretty sure the drag force is proportional to $v^2$. This means a $45^{\circ}$ angle should halve the terminal velocity. So, while I think you'd be safe saying the max speed is half the terminal velocity you looked up, I'm not willing to bet your rider's safety on that. That said, I know that whatever the real equations say, the actual terminal velocity of your rider will most definitely be less than the terminal velocity of a falling person. Go back through all the reasoning in this paragraph and you'll see we've concluded that the maximum velocity of your rider is most definitely less than the minimum of the two numbers you just looked up.

And if you want to throw friction back in at any point, then the max velocity drops even further (I'd also feel comfortable telling you to use half the terminal velocity of a person instead of the full velocity).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.