Answer:
The ratio of mass that is discarded is determined by this equation:
M - m = (3a/2)/(g²- (a²/2) - (ag/2))
Explanation:
The force acting on an object in motion is defined by the equation:
F = ma
Additionally, there is a gravitational force consistently acting downwards on the object, defined as g = 9.8 ms⁻²
For convenience, we will utilize a positive notation for downward acceleration and a negative notation for upward acceleration.
Case 1:
The hot air balloon has mass = M
Acceleration = a
Upward thrust from hot air = F = constant
Gravitational force acting downward = Mg
The net force on the balloon can be expressed as:
Ma = Gravitational force - Upward Force
Ma = Mg - F (since the balloon moves downward, that means Mg > F)
F = Mg - Ma
F = M (g-a)
M = F/(g-a)
Case 2:
After releasing the ballast, the new mass becomes m. The new upward acceleration is -a/2:
The net force is expressed as:
-m(a/2) = mg - F (The balloon is moving upwards, hence F > mg)
F = mg + m(a/2)
F = m(g + (a/2))
m = F/(g + (a/2))
Determining the fraction of the mass initially dropped:
![M-m = \frac{F}{g-a} - \frac{F}{g+\frac{a}{2} }\\M-m = F*[\frac{1}{g-a} - \frac{1}{g+\frac{a}{2} }]\\M-m = F*[\frac{(g+(a/2)) - (g-a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{g+(a/2) - g + a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{(3a/2)}{g^{2}-\frac{a^{2}}{2}-\frac{ag}{2}} ]](https://tex.z-dn.net/?f=M-m%20%3D%20%5Cfrac%7BF%7D%7Bg-a%7D%20-%20%5Cfrac%7BF%7D%7Bg%2B%5Cfrac%7Ba%7D%7B2%7D%20%7D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7B1%7D%7Bg-a%7D%20-%20%5Cfrac%7B1%7D%7Bg%2B%5Cfrac%7Ba%7D%7B2%7D%20%7D%5D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7B%28g%2B%28a%2F2%29%29%20-%20%28g-a%29%7D%7B%28g-a%29%28g%2B%28a%2F2%29%29%7D%20%5D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7Bg%2B%28a%2F2%29%20-%20g%20%2B%20a%29%7D%7B%28g-a%29%28g%2B%28a%2F2%29%29%7D%20%5D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7B%283a%2F2%29%7D%7Bg%5E%7B2%7D-%5Cfrac%7Ba%5E%7B2%7D%7D%7B2%7D-%5Cfrac%7Bag%7D%7B2%7D%7D%20%5D)