A trolley runs along a straight, level floor at a steady 3.0 m/s to the right.

Step 1. Velocity over the ground. Relative to the trolley, the ball moves backwards at $10 \times 0.6 = 6\ \text{m/s}$ and upwards at $10 \times 0.8 = 8\ \text{m/s}$. The trolley carries it forwards at $3.0\ \text{m/s}$. So over the ground the ball moves backwards at $6\ \text{m/s} - 3.0\ \text{m/s} = 3.0\ \text{m/s}$, and upwards at $8\ \text{m/s}$:

The ball still moves backwards, but much more steeply than the gun points.

Step 2. Greatest height. Only the upward part matters: the ball rises a further $\dfrac{(8\ \text{m/s})^2}{2 \times 10\ \text{m/s}^2} = 3.2\ \text{m}$ above the muzzle, so its greatest height is $1.0\ \text{m} + 3.2\ \text{m} = 4.2\ \text{m}$ above the floor.

Step 3. Time of flight. Take upwards as positive. The ball ends $1.0\ \text{m}$ below its start: $-1 = 8t - 5t^2$ (in metres and seconds), so $5t^2 - 8t - 1 = 0$ and

Step 4. Where it lands. Over the ground the ball moves backwards at $3.0\ \text{m/s}$ the whole time, so it lands $3.0\ \text{m/s} \times 1.72\ \text{s} = 5.15\ \text{m}$ behind the point below the muzzle at firing. Meanwhile the trolley moves $3.0\ \text{m/s} \times 1.72\ \text{s} = 5.15\ \text{m}$ forwards. So the ball is $5.15\ \text{m} + 5.15\ \text{m} = 10.3\ \text{m}$ behind the muzzle.

Step 5. Straight up and down. The ball's horizontal velocity over the ground is $3.0\ \text{m/s} - 10\cos\theta$. It is zero when $\cos\theta = \dfrac{3.0}{10} = 0.3$, that is $\theta = 72.5^\circ$.