Two players practise a pass on a level field…

Two players practise a pass on a level field. The figure shows the view from above. The passer P stands $24\ \text{m}$ from a straight touchline, and O is the point of the touchline nearest to P. A winger W runs along the touchline at a steady $6.0\ \text{m/s}$ . P kicks the ball along the ground. Take the ball to roll in a straight line at a steady speed (ignore the slowing by the grass), and treat the players as points.

Top view of a field. A straight touchline runs across the top. The passer P stands 24 m below the point O of the touchline. The winger W runs to the right along the touchline at 6.0 m/s. In move (a) W is at O; in move (b) W starts 18 m to the left of O.

At the moment of the kick, W is at O. The ball rolls at $10\ \text{m/s}$ . At what angle to the line PO must P kick it so that it reaches W? How long does the ball roll, and how far does W run in that time?
In another move, W is $18\ \text{m}$ before O at the moment of the kick, running towards O and on past it. What is the least ball speed with which the pass can still reach W? With that speed, how long does the ball roll, and how far past O does W receive it?
In the move of (b), the ball rolls at $5.0\ \text{m/s}$ . There are now two directions in which P can kick it to reach W. For the quicker of the two passes, find the rolling time and the angle between the ball's path and the touchline.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Use the winger's frame. In it, W stands still and the ball must travel straight along the line from P to W. Its velocity there is $\vec u - \vec v_W$ , where $\vec u$ is its velocity over the ground. So $\vec u = \vec v_W + \vec w$ , with $\vec w$ along the line from P to W's starting point. The ball arrives after a time (distance PW) $\div\ |\vec w|$ .

Step 1. Part (a). Here PW is the line PO, at right angles to the touchline, and $\vec v_W$ runs along the touchline. So $\vec u$ is the long side of a right-angled triangle with sides $\vec v_W$ and $\vec w$ . The ball's sideways part must match W's speed:

\displaystyle u\sin\theta = v_W, \qquad \sin\theta = \frac{6.0}{10} = 0.6, \qquad \theta = 36.9^\circ

The part along PO is $|\vec w| = 10 \times 0.8 = 8.0\ \text{m/s}$ , so the ball rolls for $\dfrac{24\ \text{m}}{8.0\ \text{m/s}} = 3.0\ \text{s}$ . In that time W runs $6.0\ \text{m/s} \times 3.0\ \text{s} = 18.0\ \text{m}$ .

Step 2. The geometry of (b). Now $PW = \sqrt{18^2 + 24^2}\ \text{m} = 30\ \text{m}$ . W's velocity makes an angle $\varphi$ with the line from W back to P, where $\cos\varphi = \dfrac{18}{30} = 0.6$ and $\sin\varphi = 0.8$ . Split $\vec v_W$ into a part along WP, $3.6\ \text{m/s}$ towards P, and a part at right angles to PW, $4.8\ \text{m/s}$ .

Step 3. The least speed. In $\vec u = \vec v_W + \vec w$ , the vector $\vec w$ can have any length but only the direction from P to W. The ball must at least cancel the part of $\vec v_W$ at right angles to PW. So the least speed is

\displaystyle u_{\min} = v_W\sin\varphi = 6.0\ \text{m/s} \times 0.8 = 4.8\ \text{m/s}

Then the ball is kicked at right angles to PW, and $\vec w$ just cancels the part of $\vec v_W$ along WP: $|\vec w| = 3.6\ \text{m/s}$ . The gap of $30\ \text{m}$ closes at $3.6\ \text{m/s}$ , so $t = \dfrac{30\ \text{m}}{3.6\ \text{m/s}} = 8.33\ \text{s}$ . W runs $6.0\ \text{m/s} \times 8.33\ \text{s} = 18\ \text{m} + 32\ \text{m}$ , so W receives the ball $32\ \text{m}$ past O. A ball slower than the winger can still reach W, because W runs towards the place where the ball will be.

Step 4. Part (c). Let $|\vec w| = k$ . Along PW and at right angles to it, $\vec u$ has the parts $k - 3.6\ \text{m/s}$ and $4.8\ \text{m/s}$ . Its size must be $5.0\ \text{m/s}$ :

\displaystyle (k - 3.6)^2 + 4.8^2 = 5.0^2 \quad\Rightarrow\quad k = 3.6 \pm 1.4,\ \text{so}\ k = 5.0\ \text{m/s}\ \text{or}\ 2.2\ \text{m/s}

The rolling times are $\dfrac{30\ \text{m}}{k}$ : $6.0\ \text{s}$ or $13.64\ \text{s}$ . For the quicker pass, $\vec w$ has the parts $3.0\ \text{m/s}$ backwards along the touchline and $4.0\ \text{m/s}$ towards it. Adding $\vec v_W$ , the ball's velocity has the parts $3.0\ \text{m/s}$ along the touchline and $4.0\ \text{m/s}$ towards it, so its path makes $\tan^{-1}\dfrac{4.0}{3.0} = 53.1^\circ$ with the touchline.

Answer: (a) $36.9^\circ$ to PO, towards W's run; $3.0\ \text{s}$ ; $18.0\ \text{m}$ . (b) $4.8\ \text{m/s}$ , kicked at right angles to PW; $8.33\ \text{s}$ ; $32\ \text{m}$ past O. (c) $6.0\ \text{s}$ , along a path at $53.1^\circ$ to the touchline.

Check. In (c), after $6.0\ \text{s}$ the ball has moved $4.0 \times 6.0 = 24\ \text{m}$ towards the touchline and $3.0 \times 6.0 = 18\ \text{m}$ along it, so it reaches the touchline $18\ \text{m}$ past O. W has run $6.0 \times 6.0 = 18 + 18\ \text{m}$ from $18\ \text{m}$ before O, which also puts W $18\ \text{m}$ past O.

Watch out. Do not aim at where the winger is. Aim so that, in the winger's frame, the ball comes straight at the winger. The time of flight is the distance PW divided by the closing speed along PW.

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The worked solution

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