Let's solve the questions in this problem step-by-step.

Given:

• Time of fall, $t = 0.5 \, \text{s}$
• Acceleration due to gravity, $g = 10 \, \text{m/s}^2$

(i) What is its speed on striking the ground?

The speed of an object in free fall from rest (initial velocity $u = 0$) is given by: $v = u + gt$ Since $u = 0$, this simplifies to: $v = gt = 10 \times 0.5 = 5 \, \text{m/s}$

So, the speed on striking the ground is 5 m/s.

(ii) What is its average speed during the 0.5 s?

The average speed for an object in free fall from rest can be calculated as the average of the initial and final speeds: $\text{Average speed} = \frac{u + v}{2}$ Substitute $u = 0$ and $v = 5 \, \text{m/s}$: $\text{Average speed} = \frac{0 + 5}{2} = 2.5 \, \text{m/s}$

So, the average speed during the 0.5 s is 2.5 m/s.

(iii) How high is the ledge from the ground?

The distance fallen (height) is given by the formula: $s = ut + \frac{1}{2}gt^2$ Since $u = 0$, this simplifies to: $s = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times (0.5)^2$ Calculating this: $s = \frac{1}{2} \times 10 \times 0.25 = 1.25 \, \text{m}$

So, the height of the ledge from the ground is 1.25 m.

Would you like a more detailed breakdown, or do you have further questions?

Here are some related questions to deepen understanding:

1. What would be the final speed if the car fell for 1 second instead of 0.5 seconds?
2. How would the average speed change if the time duration were doubled?
3. What would be the effect on the final speed if $g$ were 9.8 m/s² instead of 10 m/s²?
4. How high would the ledge be if the car took 1 second to fall?
5. How can we calculate the time it would take for the car to fall from a height of 5 meters?

Tip: In free fall problems, understanding the equations of motion is essential as they help solve questions related to speed, distance, and time easily.