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I was doing a physics problem, and it seemed simple at first, but now it doesn't at all, and I need help. Here it is.

Three identical spheres are thrown from the same height above the ground. Sphere X is thrown vertically up, sphere Y is thrown horizontally, and sphere Z is thrown vertically down. All three spheres are thrown with the same speed. Air resistance is negligible. Which sphere or spheres initially (because they bounce, so it means the first bounce) collide with the ground with the greatest speed.

So all 3 spheres are height h above the ground, and thrown at the same speed v. X is thrown straight up, Y is thrown straight right, and Z is thrown straight down.

So I went about doing it normally, first noting that X will go up and then come back down with the same exact velocity as Z. So I thought it was simple, X and Z both hit the ground with the greatest speed. Then I looked at the answer, and I had gotten it wrong; turned out the answer said that all of them hit the ground at the same speed. So I thought about it and realized I had just forgotten that horizontal speed counts too; Y is thrown to the side and that velocity adds on to the speed, and so it does hit the ground with the same speed. I felt satisfied that I had learned where my mistake was. Fast forward a day, and I randomly start thinking about it again. Then as I think about it I realize that Y can't hit the ground at the same speed, because of the Pythagorean theorem. My reasoning was that sphere Y will have 2 separate components of it's velocity when it hits the ground: It will have a velocity v to the right, and a velocity $v_h$ that it gained from falling. (Which should be $\sqrt{2gh}$, if that matters). So v points to the right, and $v_h$ points down. To get the total speed of sphere Y we should use the Pythagorean theorem and get $$v_y = \sqrt{v^2 + v_h^2}$$ So I worked this out and I thought, "Oh, the answer must be wrong!" But I kept looking, because I didn't think they would be wrong and me right; what are the odds? So then I tried to solve it using energy, and I got their answer! Because at the beginning all of the spheres have potential energy $U_s = m_sgh$, and kinetic energy $K_s=\frac{1}{2}m_sv^2$, where $m_s$ is the mass of the spheres. This means that their energies $E_{sx}$, $E_{sy}$, and $E_{sz}$ must all be the same At the end, when the spheres are about to bounce off of the ground, their potential energies must all be zero because they are at the ground, so there is no gravitational potential energy. Which means that their kinetic energy must all be the same, as there were no non-conservative forces involved! So I'm getting contradictory answers! Kinematics says Y has less velocity, and energy says it has the same velocity! I am very confused, and I'm sure I must've made a mistake somewhere, so if somebody could please help me I would really appreciate it.

If that was a bit confusing and you need any clarification, just ask me. Also, this is my first question on SO, so if I did anything wrong feel free to tell me :)

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closed as off-topic by David Z May 4 at 11:00

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    $\begingroup$ Why exactly do you think that "Y can't hit the ground at the same speed, because of the Pythagorean theorem"? Yes, Y's velocity would have two components: horizontal and vertical, so what? $\endgroup$ – lesnik May 3 at 15:46
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    $\begingroup$ @lesnik Because to get ||x + y|| you can't just add ||x|| and ||y||. For example, if vector x is pointing straight up with a magnitude of 2, and y is pointing straight down with a magnitude of one, what is ||x + y||? It's not 3, like you might expect. This is a bit more subtle, but it's the same concept. $\endgroup$ – PSCoder May 3 at 16:09
  • $\begingroup$ I am going to ask a dumb question, but wouldn't they all hit at the same speed (if they were high enough off the ground to) because of the acceleration of gravity here on earth. Once they hit terminal velocity, they cannot go faster.... $\endgroup$ – Rick May 3 at 16:54
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    $\begingroup$ Lol, it does say no air resistance :P $\endgroup$ – PSCoder May 3 at 17:06
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For X or Z: Final speed would be:$\sqrt{v^2+2gH}$(Using equation of motion for a uniformly accelerated motion). For Y , you saw, the net speed would be the resultant of horizontal and vertical components: $\sqrt{v^2+(\sqrt{2gH})^2}$.(horizontal component is v, and vertical is $\sqrt{2gH}$. So, kinematics and energy conservation don't contradict! Good efforts, though.

Note: you would have observed, how energy conservation is more convenient to use in such situations. Makes life easier!

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  • $\begingroup$ Yes, I did notice that! But how did you get sqrt(v^2 + 2gh) for X and Z, won't the sqrt(2gh) just add onto the v? $\endgroup$ – PSCoder May 3 at 16:06
  • $\begingroup$ The equation for uniformly accelerated motion is $v_f^2=v_i^2+2aS$. $\endgroup$ – Tojrah May 3 at 16:07
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    $\begingroup$ OH!!! This makes so much sense now! Thank you so much! I was assuming that ball Y would gain the same velocity from falling as X and Z, but forgot that because X and Z are heading down already they get down faster and so don't accelerate as fast! Thank you, somehow I didn't know about this equation :) $\endgroup$ – PSCoder May 3 at 16:14
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    $\begingroup$ @PSCoder - they do accelerate as fast. Exactly as fast... $\endgroup$ – Rory Alsop May 3 at 16:50
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    $\begingroup$ @RoryAlsop Whoops, sorry about that! I meant they don't accelerate as much $\endgroup$ – PSCoder May 3 at 17:07
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Your logic seems to be the following:

Suppose that the velocity that a ball would reach if it were just dropped is $v_1$. Suppose that it's instead thrown with velocity $v_2$. The velocity of the horizontally thrown ball will be $\sqrt{v_1^2+v_2^2}$, but the velocity of the vertically thrown ball will be $v_1+v_2$. Since $v_1+v_2 \neq \sqrt{v_1^2+v_2^2}$, we have a contradiction.

There are a few problems with that.

One is that you didn't explicitly state your claim that the vertically thrown ball will have velocity $v_1+v_2$. It's important to make your reasoning as explicit as possible, not only so other people can follow, but also because it makes it more likely that you yourself will spot your error.

The other problem is that this claim is wrong. There are several ways of seeing this:

-Momentum equation Impulse, or change in momentum, is equal to $force \times time$. The ball thrown downwards will be in the air less time than the dropped ball, so the gravitational force will be acting for less time. Thus, the impulse imparted by gravity to the thrown ball will be less than the impulse imparted to the dropped ball, and thus the change in velocity will be smaller.

-Conservation of energy The dropped ball starts out with an energy of $PE = mgh$, and ends with an energy of $KE = \frac 12 mv_1^2$. So $mgh = \frac 12 mv_1^2$. The thrown ball starts out with an energy of $KE+PE = \frac 12 mv_2^2+mgh$. We can then substitute $mgh = \frac 12 mv_1^2$ into that equation and get the the thrown ball starts out withh $KE+PE = \frac 12 mv_2^2+\frac 12 mv_1^2$. If its final velocity is $v_3$, then by conservation of energy we have $\frac 12 mv_3^2 = \frac 12 mv_2^2+\frac 12 mv_1^2$. Canceling out the $\frac12 m$, we get that $v_3^2 = \ v_2^2+v_1^2$, or $v_3= \sqrt{v_1^2+v_2^2}$, which is the same as for the horizontally thrown ball.

-Kinematics equations You should be familiar with the formula that for constant acceleration, $v_{final}^2-v_{initial}^2=2ad$. This can be derived from conservation of energy similar as above, or by using calculus. This gives $v_{final}^2=v_{initial}^2+other stuff$ rather than $v_{final}=v_{initial}+other stuff$, as you seem to be assuming.

So, the takeaway is that often, it's not velocities, but the squares of velocities, that add.

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  • $\begingroup$ Ok, I'll try to be clearer next time! I did indeed discover the problem after I saw Tojrah's answer, I can't believe I missed that X and Z would be in the air for less time! Sometimes it's the little things you miss :) Thank you! $\endgroup$ – PSCoder May 3 at 17:23
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I'm quite sure that all spheres will hit the ground at the same speed.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z May 7 at 9:07
  • $\begingroup$ @David Z I wanted to comment about my chat with hyde. $\endgroup$ – James Montagne May 10 at 14:17
  • $\begingroup$ @David Z I may have made a disparaging remark, after hyde lectured me on the nature of physics, but I immediately took it back. At any rate, usually when someone wants to end a conversation they give you some sort of a hint. Not just ghost on out. $\endgroup$ – James Montagne May 10 at 14:45
  • $\begingroup$ James, I'm not sure what you're getting at, but if you want to continue the chat, please do so in the chat room. (Also why bring up disparaging remarks? If that did happen, I don't see how it's related to what's going on here.) $\endgroup$ – David Z May 10 at 19:35
  • $\begingroup$ @David Z I'm sorry, I'm new to the physics Exchange and did not know how to address you without it being a continuation of the chat, which seems to have been discontinued by the other party, without them letting me know that they were done. It kind of left me out of sorts. $\endgroup$ – James Montagne May 10 at 21:38

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