Define the term free fall - NSC Physical Sciences - Question 3 - 2018 - Paper 1

Given that the object is in free fall, we can use the kinematic equation:

$$ext{Displacement} (s) = ut + \frac{1}{2} a t^2$$

where:

• Initial velocity (u) = 0 (since the ball is dropped),
• Acceleration (a) = 9.8 m/s² (acceleration due to gravity),
• Time (t) = 1 s.

Substituting we have:

$$\text{s} = 0 imes 1 + \frac{1}{2} \times 9.8 \times (1)^2 = 4.9 m$$

The total height of point A above the ground is therefore:

\text{Height} = 2s = 2(4.9) = 9.8 m $$.

To find the velocity of the ball just before it strikes the ground, we can again use the kinematic equation:

$$v^2 = u^2 + 2as$$

Where:

• v = final velocity,
• a = acceleration due to gravity (9.8 m/s²),
• s = height (9.8 m).
• u = 0 (initial velocity).

Substituting:

$$v^2 = 0 + 2(9.8)(9.8)$$

This simplifies to:

v^2 = 192.08$$

v = \sqrt{192.08} \approx 13.86 ext{ m/s}

$$$$

To find the average net force, we can use Newton’s second law:

$$F = ma$$

Where:

• m = mass of the ball (0.4 kg),
• a = average acceleration during the brief contact with the ground.

To find a, we consider the change in velocity:

Change in velocity = Final velocity - Initial velocity = 0 - (-13.86)

So:

Average acceleration =

$$a = \frac{\Delta v}{\Delta t} = \frac{13.86}{0.2} \approx 69.3 m/s²$$

Now we can calculate the force:

F = 0.4 imes (69.3 + 9.8) \approx 0.4 \times 79.1 \approx 31.64 N $$.