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Ball as a .

During its projectile motion toward the player, the ball is subject to an external influence from the earth(gravity). During the collision, we assume that the force from the player's head (contact force) is much larger than gravity.

Projectile Motion (One-Dimensional Motion with Constant Velocity in the y direction and One-Dimensional Motion with Constant Acceleration in the z direction) plus .

Although we are asked for the impulse acting on the player's head, it is simpler to calculate the impulse delivered to the ball by the player's head and then find the desired quantity using Newton's 3rd Law.

\begin{large} \[ t = \frac{y_{f} - y_{i}}{v_{y}} = \frac{y_{f}}{v_{y}} \] \end{large}

\begin{large}\[ z_{f} = z_{i} + v_{z,i}t + \frac{1}{2}a_{z}t^{2} \]\end{large}

\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}

\begin{large}\[ z_{f} = y_{f}+\frac{1}{2}a_{z}\frac{y_{f}^{2}}{v_{y}^{2}} \] \end{large}

\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}

\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}

\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}

Can you prove to yourself that the minus sign is the only consistent choice when taking the square root above?

\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}

\begin{large}\[ v_{i} = \sqrt{v_{x,i}^{2}+v_{y,i}^{2}+v_{z,i}^{2}} = \mbox{13.3 m/s}\]\end{large}

\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}

\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}

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.

Technically, we have not found I_bh, 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 I_bh by including the effects of gravity assuming a collision time of 0.050 s?

\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}

Player's head as a .

We are only interested in the influence from the soccer ball (collision force).

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