A

(i) Where does the ball hit the ground?
To determine where the ball hits the ground, we start by calculating the time it takes for the ball to reach the ground. The ball is thrown horizontally at a speed of 12 m/s from a height of 10 m. Since the vertical motion is independent of the horizontal motion, we can use the equation for vertical motion:
$$h = \frac{1}{2} g t^2$$
where h is the height (10 m), g is the acceleration due to gravity (approximately 9.81 m/s²), and t is the time in seconds. Rearranging this gives:
$$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 10}{9.81}} \approx 1.43 s$$
During this time, the horizontal distance covered by the ball, which has an initial horizontal velocity of 12 m/s, is given by:
$$\text{Distance} = \text{velocity} \times \text{time} = 12 \times t = 12 \times 1.43 \approx 17.16 m$$
Therefore, the ball hits the ground approximately 17.16 m horizontally from the point it was thrown.

(ii) How long after the ball hits the ground does the platform reach ground level?
The platform is descending at a constant speed of 3 m/s. To find the time it takes for the platform to reach the ground after the ball has hit the ground, we can use the same height of 10 m to find the time taken by the platform:
$$t_{platform} = \frac{h}{\text{velocity}} = \frac{10}{3} \approx 3.33 s$$
Since the ball takes approximately 1.43 seconds to hit the ground, the time after the ball hits the ground until the platform reaches the ground is:
$$3.33 - 1.43 = 1.90 s$$

(iii) With what velocity does the ball hit the ground?
The ball has both horizontal and vertical components of velocity when it hits the ground. The horizontal component is 12 m/s, and the vertical component can be calculated using:
$$v_y = g imes t \approx 9.81 \times 1.43 \approx 14.03 m/s$$
To find the resultant velocity of the ball upon impact, we can use the Pythagorean theorem:
$$v = \sqrt{v_x^2 + v_y^2} = \sqrt{12^2 + 14.03^2} \approx \sqrt{144 + 196.84} \approx \sqrt{340.84} \approx 18.46 m/s$$
Therefore, the ball hits the ground with a velocity of approximately 18.46 m/s.