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Say we throw a ball from a height of $10$ m, $10^\circ$, at $12$ m/s.

The $x$-component of the velocity vector is $12\cos(10^\circ)$ m/s and the $y$-component is $12\sin(10^\circ)$ m/s. From the kinematic equations, we get

$$x=12\cos(10^\circ)t$$

$$y=-0.5(9.8)t^2+12\sin(10^\circ)t+10$$

Assuming the coefficient of restitution is $0.7$. The second bounce has a velocity of $12(0.7)$. We solve for the initial points of the second bounce

\begin{align} -0.5(9.8)t^2+12\sin(10)t+10&=0\\ -4.9t^2+12\sin(10)t+10&=0\\\end{align} \begin{align}t&=\frac{-12\sin(10)+\sqrt{(-12\sin(10))^2-4*(-4.9)*10}}{-9.8}\\ t&\approx1.46115,-2.79344\\ x'&=12\sin(10)(1.46115) \end{align}

The parameteric equation of the second bounce is

$$x_1=12\cos(10^\circ)t+x(x')$$

$$y_1=(0.7)12\sin(10^\circ)t-(0.5)(9.8)t^2$$

Is this correct. If not why so?

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  • $\begingroup$ You seem to have dropped the coefficient of restitution term from your final equations. $\endgroup$ – Acccumulation Dec 11 '17 at 3:23
  • $\begingroup$ I assume the coefficient of restitution is the quotient between the final velocity of the first bounce and the initial velocity of the second bounce. You have applied it to the acceleration of velocity which remains unchanged. The position for the bounce is the same before and after the bounce, you just have to find the initial velocity so the coefficient of restitution is satisfied (It may affect the x velocity or not, you'll know) $\endgroup$ – ebabio Dec 11 '17 at 3:28
  • $\begingroup$ @Acccumulation The coefficient 0.7 was included in the final equations. $\endgroup$ – Arbuja Dec 11 '17 at 4:00
  • $\begingroup$ @ebabio I edited my original answer. I applied the coefficient of restitution to the y-component of the velocity. $\endgroup$ – Arbuja Dec 11 '17 at 12:48
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A couple of minor errors/inaccuracies here and there, such as the factor $\sin(10^\circ)$ in your expression for $x'$ which should actually be $\cos(10^\circ)$; the notation $x(x')$ in the final equations is unclear (that should be simply $x'$); in your final equations, you seem to apply the restitution coefficient only to the y-component, and that may be intentional - physically, the deformation would only affect motion normal to the surface - but is inconsistent with your earlier claim that "the second bounce has a velocity of $12(0.7)$." (And while I'm nitpicking, this is speed, not velocity, and should have the qualification "initial" before it).

But there is one big mistake, and it is exactly in that premise I just quoted. The initial velocity after the bounce can be figured out from the velocity just before the bounce, using the restitution coefficient - that is correct; but you assume that the speed just before the bounce would be $12$, and that is not correct, because the bounce happens at a point lower than the launch, so it is not symmetric with it. You can find the velocity components just before the bounce by substituting the time you found into the velocity equations, $v_x = 12\cos(10^\circ)$, $v_y = 12\sin(10^\circ) - 9.8t$.
You will see that the $v_y$ will not be $-12\sin(10^\circ)$ as you implicitly assumed.

Another way to find the final speed before the bounce is from conservation of energy, $12^2 + 2(9.8)(10) = v^2$; this will clearly produce a value greater than $12$. You can then use this to find $v_y$ because you know $v_x = 12\cos(10^\circ)$ throughout, and $v_x^2 + v_y^2 = v^2$. Then apply the restitution factor as intended (presumably, only on the y-component).

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  • $\begingroup$ Thank You, but I get different results when substituting the time of the first bounce into $v_x=12\cos(10)$ and $v_y=12\sin(10)-9.8t$ vs. solving for $v_y$ using $v_x^2+v_y^2=v^2$ $\endgroup$ – Arbuja Dec 11 '17 at 14:29
  • $\begingroup$ the time is wrong, don't know why, but check your calculations: Wolfram gives me $t \approx 1.65694$. @Arbuja $\endgroup$ – Nick Pavlov Dec 11 '17 at 16:05
  • $\begingroup$ Your right. However, the $v_y=12\sin(10)-9.8t$ vs $v_x^2+v_y^2=v^2$ gives a different answer. $\endgroup$ – Arbuja Dec 11 '17 at 16:16
  • $\begingroup$ It should give the same results when the correct time is used. If you are claiming that they are not, I would suggest posting the calculations that you think demonstrate that, @Arbuja $\endgroup$ – Nick Pavlov Dec 11 '17 at 16:21
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    $\begingroup$ that should be $(12^2 + 2\times9.8\times1.65694)$ in the first term under the root and should not be squared (it already is $v^2$) @Arbuja $\endgroup$ – Nick Pavlov Dec 11 '17 at 17:29

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