# A hot air balloon is rising straight upward with a constant speed of 8.80 m/s. When the basket of the balloon is 100.0 m about the ground a bag of sand tied to the basket comes loose. What is the greatest height of the bag of sand during its fall to the ground? • When the balloon goes upward with a velocity of 8.8m/s all items inside it has the same velocity of 8.8m/s.
• When the sand bag get loosed it also have that speed at the moment of loose.
• The balloon goes upwards because it has a force upward because of hot air.
• So all the parts inside the balloon is subjected to this force.
• When the sand bag get loosened it will loss the connection with the balloon.
• So this force due to hot air will no longer apply on sand bag.
• Due to the sand bag's weight it should come down.
• Since it has a velocity of 8.8m/s at the moment of loose it will go some distance upward under gravity and then start to come down.

Using equation of motion;

uarrv^2 = u^2+2as

v = 0 since at the greatest height it suddenly get to rest.

u = 8.8m/s

g = -9.8m/s^2 Since the bag travels under gravity

s = ?

uarrv^2 = u^2+2as

0 = 8.8^2-2xx9.81xxs

s = 3.95m

So the height from ground is 100+3.95 = 103.95m.

Assumption

There is no air resistance applied on the sand bag.