Solve Motion Problem: 50 m Drop Time & Velocity

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A hot air balloon descends at 2.0 m/s when a passenger drops a camera from 50 m above the ground. The initial calculations for the time it takes the camera to reach the ground are incorrect due to a sign convention error, as the final position should be -50 m. Adjusting the equation with the correct sign yields a time of approximately 2.96 seconds, but the user is unsure if this is accurate. It is suggested that using a more precise value for gravitational acceleration, g = 9.8 m/s², may yield a slightly better result. The discussion emphasizes the importance of correct sign conventions in motion problems.
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can someone help me with this problem? A hot air balloon is descending at a rate of 2.0 m/s when a passenger drops a camera. (a) If the camera is 50 m above the ground when it is dropped, how long does it take to reach the ground? (b) What is its velocity just before it lands? Let upward be the positive direction for this problem.

i did part a by this formula. 50 = (-2.0m/s)t + (1/2)(-10m/s2)t2
s2 = s squared t2 = t squared
i tried to solve for t and i got 2.9, but that not the right answer. anyone know what i did wrong?
 
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sign convention problem

It's a sign convention problem. Compared to where it starts to fall, the final position is -50 m, not +50 m.
 
ok so if i change the sign to -50 i get t = 2.96 that's what i had before. its not right
 
Last edited:
mikep said:
ok so if i change the sign to -50 i get t = 2.96 that's what i had before. its not right
What makes you think that your answer is incorrect? (Note that you'll get a slighty more accurate answer using g = 9.8 m/s^2.)
 

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