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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 in the y direction and One-Dimensional Motion with Constant Acceleration in the z direction) plus [Momentum and Impulse].

Approach:

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.

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:

Givens
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\begin

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[ y_

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= 0 ][z_

= 0][y_

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= \mbox

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][z_

= \mbox

Unknown macro: {2 m}

]
[ a_

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= -\mbox

Unknown macro: {9.8 m/s}

^

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] \end

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:

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\begin

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[ t = \frac{y_

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- x_{i}}{v_{y}} = \frac{y_{f}}{v_{y}} ] \end

Then, we can substitute into the equation:

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\begin

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[ z_

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= z_

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+ v_

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t + \frac

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a_

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t^

]\end

to obtain:

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\begin

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[ z_

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= \frac{v_{z,i}}{v_{y}} y_

+ \frac

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a_

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\left(\frac{y_{f}}{v_{y}}\right)^

]\end

In this equation, we can use the fact that the launch angle is 45°, which tells us vz,i = vy, so:

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\begin

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[ z_

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= y_

+\frac

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a_

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\frac{y_

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^{2}}{v_

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^{2}} ] \end

This equation is solved to obtain:

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\begin

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[ v_

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= v_

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= \sqrt{\frac{a_

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y_

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^{2}}{2(z_

-y_

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)}}]\end

We actually 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:

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\begin

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[ v_

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^

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= v_

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^

+ 2a_

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(z_

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-z_

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) = a_

\frac{y_

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^

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- 4 z_

y_

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+ 4z_

^{2}}{2(z_

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-y_

)}]\end

giving:

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\begin

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[ v_

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= \pm (y_

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-2z_

)\sqrt{\frac{a_{z}}{2(z_

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-y_

)}} ]\end

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

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