A hot-air balloon is moving vertically upwards with a constant speed of 3.00 m s-1.

Question 37

A hot-air balloon is moving vertically upwards with a constant speed of 3.00 m s^-1. A sandbag is dropped from the balloon. It takes 5.00 s for the sandbag to fall to the ground.

 

What was the height of the balloon when the sandbag was released?

A 29 m                        B 108 m                      C 123 m                      D 138 m

 

 

 

 

 

Reference: Past Exam Paper – November 2017 Paper 12 Q6

 

 

 

 

 

Solution:

Answer: B.

The balloon is moving upwards while the sandbag is falling downwards. We have motion in both the upward and downward directions.

 

Let the upward motion be positive.

 

Initial speed of sandbag: u = 3.00 m s^-1

Acceleration due to gravity: a = - 9.81 m s^-2 (downwards)

Time taken by sandbag to fall: t = 5.00 s

 

Equation of motion: s = ut + ½ at²

s = (3×5) + ½ × -9.81 × 5² = (-) 107.7 = (-) 108 m