Kinematics: Sandbag Released from Ascending Balloon

Q1. A small bag of sand is released from an ascending hot air balloon whose constant, upward velocity is 4.25 m/s. If the balloon is 220 m above the ground when the sandbag is released, how much time does it take for the sandbag to reach the ground?

Background

Topic: Kinematics in One Dimension (Vertical Motion)

This question tests your understanding of vertical motion under gravity, specifically how to analyze the motion of an object released from a moving platform. The sandbag starts with an initial upward velocity and is then influenced by gravity as it falls to the ground.

Key Terms and Formulas

• Initial velocity ($v_0$): The velocity of the sandbag at the moment it is released (upward, 4.25 m/s).

• Initial height ($y_0$): The height above the ground where the sandbag is released (220 m).

• Acceleration ($a$): The acceleration due to gravity ($a = -g = -9.8$ m/s2, downward).

• Kinematic equation for position:

  $y = y_0 + v_0 t + \frac{1}{2} a t^2$

Step-by-Step Guidance

1. Identify the known values: $y_0 = 220$ m, $v_0 = 4.25$ m/s (upward), $a = -9.8$ m/s2 (downward), and $y = 0$ m (ground level).

2. Set up the kinematic equation for vertical position: $y = y_0 + v_0 t + \frac{1}{2} a t^2$.

3. Substitute the known values into the equation: $0 = 220 + 4.25 t - 4.9 t^2$.

4. Rearrange the equation to standard quadratic form: $4.9 t^2 - 4.25 t - 220 = 0$.

5. At this point, you can use the quadratic formula to solve for $t$ (the time it takes for the sandbag to reach the ground):

   $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

   where $a = 4.9$, $b = -4.25$, and $c = -220$.

Try solving on your own before revealing the answer!

Final Answer: 6.85 s

Using the quadratic formula with $a = 4.9$, $b = -4.25$, and $c = -220$:

$t = \frac{4.25 + \sqrt{(-4.25)^2 + 4 \times 4.9 \times 220}}{2 \times 4.9} \approx 6.85$ s

The sandbag takes approximately 6.85 seconds to reach the ground after being released from the ascending balloon.