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Suppose a soccer player is taking a corner kick.  The 0.450 kg ball is launched at 45° from ground level and travels straight across the field (in the _y_ direction in the diagram) until it is contacted by an attacking player's head at a height of 2.0 m above the ground 20.0 m horizontally from the point of the kick.  After the header, the ball is travelling at the same speed as just before the header, but it is moving purely horizontally downfield (the _x_ direction in the diagram).  What is the impulse delivered to the player's head by the ball during the header?  (Ignore the effects of air resistance for this estimate.)

System:  Ball as a [point particle] subject to external influences from the earth (gravity) and the player's head (collision force).

Models:  Projectile Motion ([One-Dimensional Motion with Constant Velocity|1-D Motion (Constant Velocity)] in the _y_ direction and [One-Dimensional Motion with Constant Acceleration|1-D Motion (Constant Acceleration)] in the _z_ direction) plus [Momentum and Impulse].

Approach:  {note}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|Newton's Third Law].{note}

The problem has two parts.  First, we use the methods of projectile motion to determine the velocity of the soccer ball immediately prior to the collision.  Note that because this problem uses two horizontal coordinates, the projectile motion occurs in the _yz_ plane, with gravity in the -- _z_ direction.  Choosing the kick to originate from the point (0,0,0) at time _t_ = 0, our givens are:

{panel:title=Givens}{latex}\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}{latex}{panel}

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

Then, we can substitute into the equation:

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

to obtain:

{latex}\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}{latex}

In this equation, we can use the fact that the launch angle is 45°, which tells us _v_~z,i~ = _v_~y~, so:

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

This equation is solved to obtain:

{latex}\begin{large}\[ v_{y}= v_{z,i} = \sqrt{\frac{a_{z}y_{f}^{2}}{2(z_{f}-y_{f})}}\]\end{large}{latex}

We actually need _v_~z,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}\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}{latex}

giving:

{latex}\begin{large}\[ v_{z,f} = \pm (y_{f}-2z_{f})\sqrt{\frac{a_{z}}{2(z_{f}-y_{f})} \]\end{large}{latex}

We choose the sign that makes _v_~z,f~ negative, presuming that the ball is on the way down.