Calculating Time for an Object to Fall in One Dimension

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A hot air balloon rises at a constant speed of 2.50 m/s when a compass is accidentally dropped from it at a height of 3.00 m. The initial velocity (v_0) of the compass at the moment of release is 2.50 m/s upward, as it shares the balloon's speed. The formula h = v_0t + (1/2)at^2 should be used with appropriate signs for displacement and acceleration due to gravity (g). The discussion emphasizes the importance of understanding relative motion when calculating the time for the compass to hit the ground. Proper application of these concepts will yield the correct time for the compass's descent.
ShamTheCandle
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Hi. I am new to Physics Forums. The following problem is from Physics 6th Edition by Cutnell/Johnson.

A hot air balloon is rising upward with a constant speed of 2.50 m/s. When the balloon is 3.00 m above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?

I tried to solve this question by using formula below:

h=v_0t+\frac{1}{2}at^2

However, I am not sure what is the value of v_0. Is it zero or 2.50 m/s?

Thanks for taking your time to read. :smile:
 
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Hi

Vo will be 2.5 m/s in the direction the balloon is moving. If it wasn't then if you where in a moving car and you threw a ball it would fly to the back of the car at the speed the car is moving.. and that don't happen :)
 
If the balloonist is holding it in his hand, the instantaneous initial velocity of the compass at the time of release is zero with respect to him. Now the balloonist is moving with respect to the ground, so the initial velocity of the compass with respect to the ground is ...
 
Hi ShamTheCandle,
Welcome to PF.
Before dropping, the compass is moving up with the balloon. So its initial velocity is 2.5 m/s upward. Here displacement and g are in the down ward direction. While substituting in the formula take care about the signs
 
Thanks all! I appreciate your help. :smile:
 

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