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The question below shows the velocity of three balls projected in different directions will be the same upon hitting the ground.

Why isn't the velocity of the ball thrown horizontally the greatest?

Wouldn't this have both a horizontal velocity and a vertical, downward velocity due to gravity. Hence, wouldn't its velocity be the resultant velocity i.e. $sqrt$ (horizontal velocity$^2$ + vertical velocity$^2$), which would be greater than simply the vertical velocity of ball 1 or 2?

Following from this, the kinetic energy is said to be equal in all 3 when hitting the ground. But wouldn't the horizontally projected ball have the greatest kinetic energy, if it had the greatest speed?

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Source: https://courses.physics.illinois.edu/phys211/su2013/lectures/lecture8.pdf

EDIT: After the prompts in the comments, my calculations would go like this:

A̶l̶l̶ ̶t̶h̶r̶e̶e̶ ̶w̶o̶u̶l̶d̶ ̶h̶a̶v̶e̶ ̶t̶h̶e̶ ̶s̶a̶m̶e̶ ̶v̶e̶r̶t̶i̶c̶a̶l̶ ̶c̶o̶m̶p̶o̶n̶e̶n̶t̶ ̶o̶f̶ ̶v̶e̶l̶o̶c̶i̶t̶y̶ ̶w̶h̶e̶n̶ ̶h̶i̶t̶t̶i̶n̶g̶ ̶t̶h̶e̶ ̶g̶r̶o̶u̶n̶d̶,̶ ̶a̶s̶ ̶t̶h̶e̶i̶r̶ ̶s̶p̶e̶e̶d̶s̶ ̶w̶i̶l̶l̶ ̶a̶l̶l̶ ̶b̶e̶ ̶d̶u̶e̶ ̶t̶o̶ ̶t̶h̶e̶ ̶a̶c̶c̶e̶l̶e̶r̶a̶t̶i̶o̶n̶ ̶f̶r̶o̶m̶ ̶g̶r̶a̶v̶i̶t̶y̶.̶ (Corrected due to comments).

Only ball 3 has a horizontal component of velocity too.

So ball 1 would have velocity $Vy= u + a*t$

Ball 3 would have $Vx=u$ and $Vy=O+a*t$

So the resultant velocity for ball 3 would be $sqrt ( Vx^2 + Vy^2 ) = sqrt (u^2+(at)^2)$ which would be different (less?) than $a*t$.

Is that incorrect?

Where a = acceleration due to gravity = g, t = time, u = initial speed, x = horizontal component and y = vertical component.

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    $\begingroup$ If you calculate the 3 cases using kinematics, you'll find out the 3 have the same speed before reaching the ground. $\endgroup$ – Rick Nov 27 '17 at 19:24
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    $\begingroup$ All three balls take different time to reach the ground, so you can't just use the same $t$ in all three equations. $\endgroup$ – Jasper Nov 27 '17 at 19:46
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    $\begingroup$ -1 Not useful to the community. You are just asking us to spot the flaw in your calculation. $\endgroup$ – sammy gerbil Nov 27 '17 at 19:56
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    $\begingroup$ As a question that explains why a beginner's intuition and the calculation to support it can both be wrong and yet appear right to the beginner it's a good question. There are plenty here that could be easily answered by just reading the Wikipedia page, yet they're considered useful, it seems. $\endgroup$ – Nij Nov 28 '17 at 3:17
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    $\begingroup$ Also, they would all have the same speed only if there was no friction. As they all take different amount of time, their speeds will be reduced (by friction) differently. $\endgroup$ – Rafalon Nov 28 '17 at 8:35
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The key here is the fact that the ball going down is 'fired straight down'. In other words it isn't just dropping with the force of gravity, but with an initial velocity already given to it. Due to the SUVAT equation $v=u+at$ we can see that this initial velocity is added to what it would've been if it was just left to fall. In the answer you have given it says "They each start with the same kinetic energy". This means that each ball is fired with the same initial velocity. Let's work through this mathematically. The height that the ball fired up reaches is $$s=\frac{u^2}{2g}$$ Taking $g=10$ for simplicity, we can say that the ball reaches a height of $0.05u^2$. Using the equation $$v^2=u^2+2as$$ we can calculate it's final velocity, which comes out to be $$v^2=0+2*10*(0.05u^2+h)$$ or $$v=\sqrt{u^2+20h}$$

Let's now think about the ball fired downwards. This is going to have a final speed of $$v^2=u^2+20h$$ or $$v=\sqrt{u^2+20h}$$

Now let's do the velocity for the final ball, projected horizontally. We can calculate it's vertical velocity by using $$v^2=u^2+2ah$$ and from this we get $$v_{vert}=\sqrt{20h}$$ As there is nothing happening to the horizontal velocity it is still just $u$ at the end, so for the ball fired horizontally we have $$v_{vert}=\sqrt{20h}$$ $$v_{hor}=u$$

Here we need to calculate the magnitude of the velocity, $\|v\|$, by taking $$v=\sqrt{v_{vert}^2+v_{hor}^2}$$ We can see here that if we do this we end up with $$v=\sqrt{u^2+20h}$$

So all balls end up with the same speed at the end. It is much easier to do this sort of thing as a change in energy, however this way is also an effective method of proving it.

Hope this helps :)

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Nov 28 '17 at 8:45
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Conservation of energy provides a second way, a sanity check.

$$E = K + U$$

Allow the gravitational potential $U$ energy to be 0 on the floor and $mgh$ on the ledge.
Each ball starts with the same speed $u$. So the energy of each ball at the beginning is:

$$ \frac{1}{2} m u^2 + mgh $$

The energy when they reach the floor is:

$$ \frac{1}{2} m v^2 $$

By conservation of energy, these to quantities must be the same:

\begin{aligned} \frac{1}{2} m v^2 &= \frac{1}{2} m u^2 + mgh \\ \\ v^2 &= u^2 + 2gh \end{aligned}

So, conservation of energy not only tells us that the three balls will have the same speed when they reach the floor, but it tells us exactly what the speed will be. Crucially though, it does not tell has what the velocity is, in other words, we still don't know how fast each ball is moving horizontally or vertically. We know the length of the vector, but not its direction.

However, we can answer your question "why do all the balls have the same velocity when they reach the floor?" By noticing that they all start with the same energy, and convert the same amount of potential energy into kinetic energy.

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  • $\begingroup$ "Crucially though, it does not tell has what the velocity is" - it does in this case since we're given the initial velocity for each ball. By inspection, the velocity of the first and second ball is $-\sqrt{u^2 + 2gh}\,\hat{y}$ and the velocity of the third ball is $u\,\hat{x} - \sqrt{2gh}\,\hat{y}$ $\endgroup$ – Alfred Centauri Nov 27 '17 at 21:36
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    $\begingroup$ @AlfredCentauri What I intended is that conservation of energy alone is not enough to derive the velocities. Of course, we can obtain the velocities using the EOMs, but the whole point is that OP was having trouble with those. $\endgroup$ – Andrea Nov 27 '17 at 21:51
  • $\begingroup$ But I didn't use the EOM either. I simply looked at your last equation (derived from conservation of energy) and the given initial velocities to write the answers by inspection. $\endgroup$ – Alfred Centauri Nov 27 '17 at 22:54
  • $\begingroup$ @AlfredCentauri is "inspection" not a euphemism for EOM? +1 for an elegant answer. $\endgroup$ – LLlAMnYP Nov 28 '17 at 15:26
  • $\begingroup$ @LLlAMnYP, well of course it isn't. The last equation implies $v = \sqrt{u^2 + 2gh}$. But $v = \sqrt{v^2_x + v^2_y}$ (no EOM there). With the initial velocity components for this problem, this is all that is needed to find the velocities just before impact. $\endgroup$ – Alfred Centauri Nov 28 '17 at 15:48
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You are given four answers to choose from. Without needing to do any calculation if you examine balls 1 and 2, where ball 1 is fired straight up and ball 2 is fired straight down, and assume no other effect (air resistance, etc.), then ball 1 will have the same speed as ball 2 when it returns to its starting point. From this point both are acted on by gravity over the same distance (to the ground) and the only answer where v1 = v2 is D.

At this point, even if the question has an error, there are no alternate answers. And you could substitute some values and calculate all three cases. And you should find that the change in vertical velocity for ball 3 is different than balls 1 and 2, and that the resultant speed is the same.

And if you include simple friction from the air then when ball 1 returns to its starting point during its fall then it will have a lower speed than ball 2. From this point both are acted on by gravity over the same distance (to the ground) and ball 2 will have a higher speed than ball 1 (unless it's the case where the initial velocity is higher than the terminal velocity and the ball slows down). But you lack the information to calculate all of this and can assume the first (frictionless) case (where v1 = v2 and answer is D).

And there is always the case where the height is great enough and the effects of passing through air (instead of assuming a vacuum / no friction) result in all three balls reaching terminal velocity before they reach the ground and thus (again) v1 = v2 = v3.

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    $\begingroup$ In the limiting case $h=0$, it's easy to see that $v_2 = v_3$, as both balls are on the ground already... $\endgroup$ – yatima2975 Nov 28 '17 at 10:33

protected by Qmechanic Nov 28 '17 at 1:18

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