# Does angle of inclination between slide and the floor affect speed at the bottom? [closed]

I obtain the following question in this link.

I don't understand why the velocity at the bottom for the four slides are the same. I thought the velocity depends on the angle of inclination between slide and the floor. In other words, my answer is

$$v_D > v_A > v_B > v_C$$

Can anyone explain to me why all speeds are the same?

## closed as off-topic by knzhou, John Rennie, CuriousOne, sammy gerbil, heatherAug 4 '16 at 12:39

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – knzhou, John Rennie, CuriousOne, sammy gerbil, heather
If this question can be reworded to fit the rules in the help center, please edit the question.

• VtC voters: check the link, it is not homework. – user259412 Aug 4 '16 at 4:00
• @peterh : You misunderstand the definition of "homework" on this site. Full title is "homework and exercises," which includes all questions of a "homework-like" nature, even if done for self-study. See meta.physics.stackexchange.com/questions/714 – sammy gerbil Aug 4 '16 at 9:45
• @sammygerbil You are right. Although I think, maybe the homework (& exercises) policy could be softened on pragmatical reasons, ref1, ref2. – user259412 Aug 4 '16 at 15:57
• @peterh : I agree, the policy should be revised, although I think it should not depend on whether the question has actually been set for homework - that would be too difficult to verify. I agree this question should be on topic. But according to the official policy I think it is not. If you want to change policy, get involved on the Physics Meta site. – sammy gerbil Aug 4 '16 at 16:19

First, the reason the child is moving at all is that the potential energy at the top is converted into kinetic energy at the bottom. If the initial height is $h$ and the mass of the child is $m$, we have a final speed $v$ of $$mgh=\frac{1}{2}mv^2\to v=\sqrt{2gh}$$ This is clearly independent of the path the child takes to get to the bottom, and so given that $h$ is the same in all cases, $v$ will also be the same. The child could be moving at a non-zero initial velocity, but this doesn't affect the result; the initial energy (and thus the final energy) will still be the same.