Two players practise a pass on a level field.

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:

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

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 $5.0\ \text{m/s}$:

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.