On 16 August, 1960, Joseph Kittinger established a record for the highest altitude parachute jump - Leaving Cert Physics - Question 6 - 2012

The downward force $F$ can be calculated using Newton's second law: $F = m \cdot g_h$ where:

• $m$ is the total mass (180 kg)
• $g_h$ is the acceleration due to gravity at 31 km ($\approx 9.99 m/s²$)

Substituting the values: $F = 180 \times 9.99 \approx 1798.2 N$

To estimate the distance fallen during the first 13 seconds, we use the formula: $s = ut + \frac{1}{2} a t^2$ Assuming he started from rest, $(u=0)$: $s = 0 \cdot 13 + \frac{1}{2} \cdot 9.99 \cdot (13)^2$

Calculating: $s = \frac{1}{2} \cdot 9.99 \cdot 169 \approx 844.26 m$

Assumptions:

1. The acceleration due to gravity remains constant at this height.
2. There is no atmospheric resistance acting on Kittinger during this time.

First, convert 4 minutes and 36 seconds to seconds: $4 \text{ minutes} = 4 \times 60 = 240 \text{ seconds}$; thus, $240 + 36 = 276 \text{ seconds}$

Using the distance fallen (5 km or 5000 m) as a reference: $\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{5000}{276} \approx 18.12 m/s$

The force on a parachute depends on its area. The area $A$ of a hemispherical parachute can be calculated as: $A = \frac{1}{2} \pi r^{2}$ For a parachute of diameter 8.5 m: $r = \frac{8.5}{2} = 4.25 m\Rightarrow A_1 = \frac{1}{2} \pi (4.25)^{2} \approx 28.4 m^{2}$

For a parachute of diameter 1.8 m: $r = \frac{1.8}{2} = 0.9 m\Rightarrow A_2 = \frac{1}{2} \pi (0.9)^{2} \approx 1.27 m^{2}$

Assuming the same force applied to both, the difference in force can be calculated by: $Force_{diff} = A_1 - A_2 \approx 28.4 - 1.27 \approx 27.13 N$

At constant velocity, the upthrust $U$ equals the weight of Kittinger: $U = weight = m imes g$ Substituting values: $U = 180 kg \times 9.81 m/s^2 \approx 1766 N$