Solve Free-Fall Problem: Hot Air Balloon & Package t=2.26/2.49s

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SUMMARY

The discussion focuses on solving the free-fall problem involving a hot air balloon traveling upward at 2.2 m/s that releases a package at a height of 25 m. Two calculations for the time the package is in the air are presented: 2.26 seconds without considering the balloon's upward velocity and 2.49 seconds when accounting for the initial upward velocity of the package. The upward movement of the balloon significantly impacts the time the package remains airborne, as it retains the balloon's initial velocity upon release. The discussion emphasizes the importance of understanding initial velocity in projectile motion.

PREREQUISITES
  • Understanding of kinematic equations, specifically d = v(initial)t + 0.5(a)t²
  • Knowledge of gravitational acceleration, specifically 9.8 m/s²
  • Familiarity with the concept of initial velocity in projectile motion
  • Basic understanding of parabolic motion and frames of reference
NEXT STEPS
  • Study the effects of initial velocity on projectile motion in various contexts
  • Learn about kinematic equations in two dimensions
  • Explore examples of projectile motion in real-world scenarios, such as bombs dropped from aircraft
  • Watch educational videos on projectile motion and reference frames for deeper understanding
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding the principles of projectile motion and the effects of initial velocity on falling objects.

chemguy990
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Homework Statement


hot air balloon traveling vertically upward at 2.2 m/s. at 25 m above ground, releases a package. Solve for t.


Homework Equations


d=v(initial)t + 0.5(a)t2


The Attempt at a Solution



25m = [0.0 m/s(t)] + ½(9.8m/s2)t2
25/4.9 = t2
t=2.26 seconds

My question is, does the upward movement of the balloon impact the time that the package is in the air? If so, would it be:

25 = -2.2t +0.5(9.8)t2
t= 2.49 seconds
 
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chemguy990 said:
My question is, does the upward movement of the balloon impact the time that the package is in the air? If so, would it be:

25 = -2.2t +0.5(9.8)t2
t= 2.49 seconds
Absolutely. The fact that the package was riding with the balloon tells you its initial velocity.
 
Yes, (your second attempt, although I didn't check the arithmetic)
The ball would initially be moving at the same velocity as the air balloon all though gravity would immediately cause it to slow down to a stop, then to speed up on its way down. For some examples to help you get this concept down, consider a jet dropping a bomb - the bomb initially has the same forward velocity as the jet - it doesn't simply fall straight down from the point of release. Instead, it follows a parabolic path (ignoring air resistance and the curvature of the earth.) Another example would be to drop a dense object out the window of a moving car (at not too high of a speed such that air resistance becomes a huge factor.) The object will remain beside the drivers door while it falls to the ground.
For another example, see this video of a cart shooting a ball vertically (with respect to the cart) - notice that the horizontal velocity of the ball is the same as the horizontal velocity of the cart at the moment it leaves the cart.
http://www.physics.umd.edu/lecdem/services/demos/demosc2/c2-25.htm

edit: (Wow, I'm slow! DocAl is too quick for me.)

What would trick people into thinking that the ball being dropped from the balloon is immediately going in the downward direction is simply because they're using the balloon as a frame of reference, rather than the ground. Plus, the low velocity of the balloon means the ball will reach a velocity of 0 m/s in only about a fifth of a second. If it were a piece separating from a rocket at a high velocity, it might be more noticeable that with respect to the ground, the piece continues moving vertically upward until it slows to a stop.
 
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