Question

A person holding a lunch bag is moving upward in a hot air balloon at a constant speed of 7.3 m/s . When the balloon is 24 m above the ground, she accidentally releases the bag. What is the speed of the bag just before it reaches the ground?

Answer 1

Explanation:

Given that,

Initial speed of the bag, u = 7.3 m/s

Height above ground, s = 24 m

We need to find the speed of the bag just before it reaches the ground. It can be calculated using third equation of motion as :

$v^2=u^2+2as$

$v^2=(7.3)^2+2\times 9.8\times 24$  

$v=\sqrt{523.69}$

v = 22.88 m/s

So, the speed of the bag just before it reaches the ground is 22.38 m/s. Hence, this is the required solution.