A hot air balloon is moving vertically - Scottish Highers Physics - Question 2 - 2023

We can use the following kinematic equation to relate the distance fallen, the initial velocity, acceleration due to gravity, and time:

$s = ut + \frac{1}{2} a t^2$
Where:

• $s = 50 \text{ m}$ (the distance fallen)
• $u$ is the initial velocity we want to find
• $a = 9.81 \text{ m/s}^2$ (acceleration due to gravity)
• $t = 3.0 \text{ s}$

Substituting the known values into the equation gives:

$50 = u(3.0) + \frac{1}{2}(9.81)(3.0)^2$

This simplifies to:
$50 = 3u + 44.145$

Rearranging for $u$ gives:

$3u = 50 - 44.145$
$3u = 5.855$
$u = \frac{5.855}{3} \approx 1.95167 \text{ m/s}$.
Since the balloon is moving upwards at the moment of release, the initial velocity is directed downwards.

Rounding the answer to the nearest whole number, the approximate initial velocity of the sandbag on release is $2.0 \text{ m/s}$ downwards. Therefore, the correct answer is B.