# Thread: Differential Equations Applications in Physics: Rising Balloon Problem

1. ## Differential Equations Applications in Physics: Rising Balloon Problem

A ball is released from a balloon at an altitude of 100 feet, The balloon at that moment is rising at the rate of 50 ft./sec.

(a) Find the distance over which the ball continues to rise.
(b) Find the time elapsed from the time the ball is dropped to the time the ball hits the ground.

2. ## Re: Differential Equations Applications in Physics: Rising Balloon Problem

Originally Posted by twilightmage13
A ball is released from a balloon at an altitude of 100 feet, The balloon at that moment is rising at the rate of 50 ft./sec.

(a) Find the distance over which the ball continues to rise.
(b) Find the time elapsed from the time the ball is dropped to the time the ball hits the ground.
What have you been able to do so far?

-Dan

3. ## Re: Differential Equations Applications in Physics: Rising Balloon Problem

Part (a)
Knowing that acceleration is equal to the gravitation acceleration, I deduced that: d²x/dt² = -g, then I integerated to find dx/dt = -gt + C.
At t=0, (dx/dt)(0) = C .: C = v0 or initial velocity.
I'm not too certain what the question for part (a) is asking since I can't imagine a ball rising once it is released...unless it is thrown upwards?
In any case, I integrated dx/dt and got x(t) = (-1/2)gt² + (v0)t + x0, then substituted values g = 9.8 m/s², v0 = 50 ft/s = 15.24 m/s,
and x0 = 100 ft = 30.48m into the equation and got: x(t) = -4.9t² + 15.24t + 30.48
but not sure if this answers part (a).

Part (b)
Assuming that the above equation for distance that I derived is correct, I simply needed to solve for t in x(t) = 0.
I used the quadratic formula and got t = 4.49s (ignoring the negative solution since time must be a positive value).

4. ## Re: Differential Equations Applications in Physics: Rising Balloon Problem

Originally Posted by twilightmage13
Part (a)
Knowing that acceleration is equal to the gravitation acceleration, I deduced that: d²x/dt² = -g, then I integerated to find dx/dt = -gt + C.
At t=0, (dx/dt)(0) = C .: C = v0 or initial velocity.
I'm not too certain what the question for part (a) is asking since I can't imagine a ball rising once it is released...unless it is thrown upwards?
The entire balloon, with the ball in it, is rising at 50 ft per second so the initial velocity, C above, is 50 ft per second. That means dx/dt= 50- ct.
Initially the balls speed is 50 feet per second upward. When does dx/dt become 0 and then negative?

In any case, I integrated dx/dt and got x(t) = (-1/2)gt² + (v0)t + x0, then substituted values g = 9.8 m/s², v0 = 50 ft/s = 15.24 m/s,
and x0 = 100 ft = 30.48m into the equation and got: x(t) = -4.9t² + 15.24t + 30.48
but not sure if this answers part (a).
No, it does't. The problem is asking for a specific time, not a formula. The ball's initial velocity is 50 m/s upward. When will it be 0? How far will it rise in that time?

Part (b)
Assuming that the above equation for distance that I derived is correct, I simply needed to solve for t in x(t) = 0.
I used the quadratic formula and got t = 4.49s (ignoring the negative solution since time must be a positive value).
I would suggest that, since height and initial speed are given in the English system, you use g= -32.2 feet per second squared rather than change everything to metric system.

5. ## Re: Differential Equations Applications in Physics: Rising Balloon Problem

Differential calculus is the study of the rates at which quantities changes. The primary objects of study in differential calculus are the derivatives of a function, related notions such as the differential and their applications.

SAT Prep

6. ## Re: Differential Equations Applications in Physics: Rising Balloon Problem

Originally Posted by nigelbranden
Differential calculus is the study of the rates at which quantities changes. The primary objects of study in differential calculus are the derivatives of a function, related notions such as the differential and their applications.

SAT Prep