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We know that if a ball rolls down a ramp, then the component of its weight acting down in the direction of the ramp is $g\sin\theta$. Therefore, we can substitute that into $s=ut+at^2/2$ to get $s=(g\sin\theta)t^2/2$, where $s$ denotes the height of the ramp.

But if you put in $s/\text{(length of ramp)}$ for $\sin\theta$, the $s$ cancels out and the equation breaks.

What am I getting wrong about this?

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  • $\begingroup$ Just curious, I've never heard term "diluted gravity", where did you find that? $\endgroup$
    – Charlie
    Commented Oct 11, 2018 at 23:49
  • $\begingroup$ @Charlie It's a famous (I think) experiment done by Galileo. Google "galileo diluted gravity" or something like that. $\endgroup$
    – Ivan
    Commented Oct 11, 2018 at 23:54
  • $\begingroup$ Oh I guess I knew it by another name since I'm not a native English speaker. $\endgroup$
    – Charlie
    Commented Oct 12, 2018 at 0:23

2 Answers 2

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where s becomes the vertical height of the ramp

This is incorrect. The displacement $s$ is the distance covered, i.e. the length of the ramp.

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The confusion arises because you're mixing variables. Let $h$ be the height of your ramp and $L$ the length, with $\theta$ being the angle with respect to the ground. Therefore, this means that:

$Sin\theta=\frac{h}{L}$

As the ball rolls down it traverses the hypotenuse, so your equation, assuming no initial velocity, should instead read:

$s=\frac{1}{2}gSin\theta t^2$

Where $s$ denotes the displacement component parallel to the hypothenuse (where the ball rolls). In other words, you used $s$ both as the distance traversed and the height of the ramp, so got that inconsistency. Now if you substitute the definition of sinus:

$s=\frac{1}{2}g(\frac{h}{L})t^2$

Which doesn't lead to any problems. So be careful with naming your variables.

As an exercise, you may want to see what happens when $s=L$. What would this mean?

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