# 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...

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?

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- 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.*

**Sources:**

When the bag of sand comes loose from the basket of the balloon, it will have two forces acting on it in mutually opposite directions: one upward due to the upward velocity of the balloon and the other downward acceleration due to gravity. As a consequence, it will rise a bit and then start falling to the ground.

For the bag of sand,

Initial upward velocity`=u`

final velocity, `v = 0` (since at the topmost point its upward velocity will become zero)

`g= 9.81` m/s^2

Applying

`v=u-g*t` (since g is downwards)

we get time to rise up, `t=(u/g)` s

Height attained in this time, in addition to the initial height of 100 m, can be obtained by applying

`s = u*(u/g)-1/2 g(u/g)^2`

`=u^2/(g) -1/2*g*(u/g)^2`

`=u^2/(g)-u^2/(2g)`

`=u^2/(2g)`

`=8.8^2/(2*9.81)` m

`=3.95` m

So, total height of the bag of sand from the ground before coming to a standstill and then beginning to fall:

`= (100+ 3.95)` m

`=103.95` m

This is the greatest height of the bag of sand during its to fall to the ground.

**Sources:**