How Far Does a Sand Bag Fall from a Rising Hot Air Balloon After Five Seconds?

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A sandbag released from a rising hot air balloon at 5.0 m/s falls a distance of 122.5 meters after five seconds. The correct calculation involves using the equation d = v_o t + 1/2at^2, which accounts for the initial upward velocity of the bag. Initially, the user mistakenly combined two separate calculations, leading to an incorrect distance of 97.5 meters. Recognizing the need to factor in the bag's initial velocity clarified the solution. Understanding the physics of motion is crucial for accurate problem-solving in this context.
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A hot air balloon is rising at 5.0 m/s when a bag is released. Five seconds later, how far below the hot air balloon is the sand bag. Answer: 122.5 m.

So I use the equation (1/2)(a)(t)^2 to figure out the distance the bag has traveled. I use the equation -Vt to figure out how far the balloon has risen. I end up getting 97.5 m when I combine the 2 but when I use just (1/2)(a)(t)^2, I get the correct answer.. Can someone please explain to me what I'm doing wrong?
 
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did you take into account that when the sandbag begins to fall it has an upward velocity of 5m/s?
 
Midy1420 said:
did you take into account that when the sandbag begins to fall it has an upward velocity of 5m/s?

Wow! I can't believe I forgot about that. Thanks a lot.
 
The actual equation is d=v_o t + 1/2at^2
 
Chi Meson said:
The actual equation is d=v_o t + 1/2at^2

Thanks!
 

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