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Hot Air Balloon Flights

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how do 4 forces of flight affect a Hot Air Balloon?

use thrust, weight, drag and lift

Hot Air Balloons aren’t exactly designed with thrust and drag in mind.
Obviously, the weight of the balloon acts upon it like any other aircraft, attempting to pull its mass to the ground. Lift is produced by the difference in density inside and out of the balloon. The Hot air within is less dense than the cool air outside, so it will attempt to rise. The mass of hot air produces enough upward lift to counteract gravity, and it pulls the entire balloon and basket assembly upwards.
A conventional hot air balloon does not have any means of producing thrust. Any forward movement is actually produced by, well, drag. Air moving around the craft will create pockets of high and low pressure which will pull the balloon around it. How much this affects the craft is entirely dependent on the air currents that it is flying in. In designing a balloon, if one wants it to move more, they would actually design it to induce more drag in the air flowing around it.


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Hotair Balloon

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Help trying to understand this question?

I dont really want the answer im trying to understand

A hotair Balloonist, rising vertically with a constant speed of 5.00m/s , releases a sandbag at the instant the Balloon is 40.0m above the ground. After it is released, the sandbag encounters no appreciable air drag.

so will 40m be the final position?
and 5m/s be the initial velocity?

First
Fup – Fdown = m a = 0 ; Fup = Fdown (v = cte)

Second: when sandbag is released

now Fup > Fdown, the balloon loses weight

Therefore the balloon starts speeding up and going up.

Up Promo Hot Air Balloon at Disneyland Park


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