3.1 Is the hot air balloon in free fall? Choose from YES or NO - NSC Physical Sciences - Question 3 - 2021 - Paper 1

To find the speed of the hot air balloon, we note that it moves upwards at a constant speed. Since stone A reaches the ground at 62.68 m/s and the balloon has been moving upwards, we establish:

Assuming no air resistance, the speed of the hot air balloon is equal to its terminal velocity when stone A is dropped, thus:

• Speed = 62.68 m/s.

Using the equation of motion: $v^2 = u^2 + 2as$ Where:

• $v = 62.68$ m/s (final velocity)
• $u = 0$ m/s (initial velocity of stone A when dropped)
• $a = 9.81$ m/s² (acceleration due to gravity)
• $s = 200$ m (distance)

Rearranging, we find time using: $s = ut + \frac{1}{2} a t^2$ Substituting and solving for t gives:

• Time it takes stone A to strike the ground = approximately 6.32 seconds.

To find the distance, we first calculate the height of the hot-air balloon at the time stone A reaches the ground:

• Height of balloon = initial height - distance fallen by stone A Using the time calculated before:
• While stone A is falling for 6.32 s, stone B is dropped 5 s later, meaning stone B has been falling for (6.32 - 5) seconds = 1.32s.
• Distance fallen by stone B = \frac{1}{2} g t^2 = \frac{1}{2} * 9.81 * (1.32^2) = approximately 8.55 m. The distance between the two at that instant is:
• Height of balloon when stone A strikes ground - distance fallen by B = 200 m - 8.55 m = 191.45 m.

In the position-time graph:

• The graph for the hot-air balloon will be a straight line moving upwards at a constant rate until the moment stone A is dropped.
• For stone A, the graph will be a curve that starts at the height of 200m and decreases non-linearly to 0m as it strikes the ground, showing the increase in speed due to gravitational acceleration. Label the balloon's graph as 'BALLOON' and stone A's graph as 'A'.