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?
- 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.`
There is no air resistance applied on the sand bag.
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
`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`
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
This is the greatest height of the bag of sand during its to fall to the ground.