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

To find the height of point A, we can use the equation of motion under gravity:

$$ext{Distance} = v_i t + \frac{1}{2} a t^2$$

Substituting for a (acceleration due to gravity, which is approximately $9.8 \, m/s^2$), $v_i = 0$, and $t = 1 \, s$:

$$ext{Distance} = 0 + \frac{1}{2} (9.8)(1^2) = 4.9 \, m$$

Since point B is halfway, the height of point A above the ground is:

$$ext{Height} = 2(4.9) = 9.8 \, m$$

Using the formula:

$$v^2 = u^2 + 2a s$$

where:

• $u = 0$ (initial velocity)
• $a = 9.8 \, m/s^2$ (acceleration due to gravity)
• $s = 9.8 \, m$ (distance fallen) Then:

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

Calculating:

$$v^2 = 192.08 \ v = \sqrt{192.08} \ v \approx 13.86 \, m/s$$

The average net force can be calculated using:

$$F_{net} = m a$$

where:

• $m = 0.4 \, kg$ (mass of the ball)
• $a = g = 9.8 \, m/s^2$ (acceleration due to gravity) Thus:

$$F_{net} = 0.4 imes 9.8 = 3.92 \, N$$

Therefore, the average net force exerted on the ball while it is in contact with the ground is approximately $3.92 \, N$.