\begin{large}\[ y_{i} = 0 \]\[z_{i} = 0\]\[y_{f} = \mbox{20 m}\]\[z_{f} = \mbox{2 m} \]
\[ a_{z} = -\mbox{9.8 m/s}^{2}\] \end{large} |
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{latex}
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The most direct way to proceed is to express the time (which we do not need to solve for) in terms of the y-velocity:| Latex |
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\begin{large} \[ t = \frac{y_{f} - y_{i}}{v_{y}} = \frac{y_{f}}{v_{y}} \] \end{large} |
Then, we can substitute into the equation: | Latex |
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\begin{large}\[ z_{f} = z_{i} + v_{z,i}t + \frac{1}{2}a_{z}t^{2} \]\end{large} |
to obtain: | Latex |
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\begin{large}\[ z_{f} = \frac{v_{z,i}}{v_{y}} y_{f} + \frac{1}{2}a_{z}\left(\frac{y_{f}}{v_{y}}\right)^{2}\]\end{large} |
In this equation, we can use the fact that the launch angle is 45°, which tells us vz,i = vy, so: | Latex |
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\begin{large}\[ z_{f} = y_{f}+\frac{1}{2}a_{z}\frac{y_{f}^{2}}{v_{y}^{2}} \] \end{large} |
This equation is solved to obtain: | Latex |
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\begin{large}\[ v_{y}= v_{z,i} =\pm \sqrt{\frac{a_{z}y_{f}^{2}}{2(z_{f}-y_{f})}} = \mbox{10.4 m/s}\]\end{large} |
We choose the plus sign, since we have set up our coordinates such that the ball will move in the + y direction. We are not finished, since we also need vz,f, the z-velocity at the end of the projectile motion and at the beginning of the ball's collision with the player's head. To find this velocity, we can use: | Latex |
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\begin{large}\[ v_{z,f}^{2} = v_{z,i}^{2} + 2a_{z}(z_{f}-z_{i}) = a_{z}\frac{y_{f}^{2} - 4 z_{f}y_{f} + 4z_{f}^{2}}{2(z_{f}-y_{f})}\]\end{large} |
giving: | Latex |
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\begin{large}\[ v_{z,f} = \pm (y_{f}-2z_{f})\sqrt{\frac{a_{z}}{2(z_{f}-y_{f})}} = - \mbox{8.35 m/s}\]\end{large} |
We choose the sign that makes vz,f negative, presuming that the ball is on the way down. | Tip |
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Can you prove to yourself that the minus sign is the only consistent choice when taking the square root above? |
Phase 2: CollisionWe have now completed the analysis of the projectile motion. Using the fact that the final velocity of the projectile motion will equal the initial velocity of the collision with the player's head, we summarize the initial velocity of the ball for the collision: | Latex |
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\begin{large}\[ v_{x,i} = \mbox{0 m/s} \]
\[ v_{y,i} = \mbox{10.4 m/s}\]
\[ v_{z,i} = \mbox{- 8.34 m/s}\]\end{large} |
The magnitude of the initial velocity is then: | Latex |
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\begin{large}\[ v_{i} = \sqrt{v_{x,i}^{2}+v_{y,i}^{2}+v_{z,i}^{2}} = \mbox{13.3 m/s}\]\end{large} |
Thus, from the information given in the problem, we will take the final velocity of the ball immediately following the collision with the player's head to be: | Latex |
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\begin{large}\[ v_{x,f} = \mbox{13.3 m/s} \]
\[ v_{y,f} = \mbox{0 m/s}\]
\[ v_{z,f} = \mbox{0 m/s}\]\end{large} |
The impulse delivered to the ball during its contact with the player's head is therefore: | Latex |
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\begin{large}\[ I_{bh} = \Delta \vec{p} = m_{ball}((v_{x,f}-v_{x,i})\hat{x}+(v_{y,f}-v_{y,i})\hat{y} + (v_{z,f}-v_{z,i})\hat{z}) = m_{ball}(v_{x,f}\hat{x} - v_{y,i}\hat{y} - v_{z,i}\hat{z}) = \mbox{6.0 kg m/s}\:\hat{x} -\mbox{4.7 kg m/s}\:\hat{y} + \mbox{3.8 kg m/s}\:\hat{z} \]\end{large} |
| Tip |
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It is important to think carefully about the expected signs when calculating a change. The ball ends up with a positive x-momentum, so the x-impulse is positive. The ball loses a positive y-momentum, so the y-impulse is negative, the ball loses a negative z-momentum, so the z-impulse is positive. |
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Technically, we have not found Ibh, but rather the total impulse on the ball during the collision. If the collision is long enough, gravity's contribution to this impulse might be non-negligible. How much of a difference would be made in the result for Ibh by including the effects of gravity assuming a collision time of 0.050 s? |
Note, however, that we were asked for the impulse delivered to the player's head. By Newton's 3rd Law, that impulse is simply: | Latex |
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\begin{large}\[ I_{hb} = -I_{bh} = -\mbox{6.0 kg m/s}\:\hat{x} +\mbox{4.7 kg m/s}\:\hat{y} - \mbox{3.8 kg m/s}\:\hat{z} \]\end{large} |
The magnitude of this impulse is 8.48 kg m/s. | 
