A light ball is thrown from ground level on a flat beach…

Step 1. Effective gravity. The acceleration has $7.5\ \text{m/s}^2$ downwind and $10\ \text{m/s}^2$ downwards. Its size is $\sqrt{7.5^2 + 10^2}\ \text{m/s}^2 = 12.5\ \text{m/s}^2$, and it points below the horizontal, downwind, at an angle whose tangent is $\dfrac{10}{7.5}$, that is $53.1^\circ$.

Step 2. Back to the hand (a). Throw the ball exactly against the effective gravity: upwind, at $53.1^\circ$ above the horizontal. Its velocity and acceleration then lie along one line, so it moves along a straight line, slows down, stops and comes back along the same line to the hand. It is like a vertical throw with $g' = 12.5\ \text{m/s}^2$:

The greatest height is $L\sin53.1^\circ = 4\ \text{m} \times 0.8 = 3.20\ \text{m}$.

Step 3. The range on level ground. Thrown at $\theta$, the ball is in the air for $T = \dfrac{2u\sin\theta}{g}$, because the wind does not change its upward motion. Downwind (plus) or upwind (minus):

Here $\dfrac{u^2}{g} = 10\ \text{m}$ and $\dfrac{a}{g} = 0.75$. The largest value of $\sin 2\theta \mp 0.75\cos 2\theta$ is $\sqrt{1 + 0.75^2} = 1.25$.

Step 4. Downwind (b). The largest value is reached when $\tan 2\theta = -\dfrac{g}{a}$, so $2\theta = 180^\circ - 53.1^\circ$ and $\theta = 63.4^\circ$. Then $R = 10\ \text{m} \times (1.25 + 0.75) = 20.0\ \text{m}$.

Step 5. Upwind (c). Now $\tan 2\theta = \dfrac{g}{a}$, so $2\theta = 53.1^\circ$ and $\theta = 26.6^\circ$. Then $R = 10\ \text{m} \times (1.25 - 0.75) = 5.0\ \text{m}$.