How long does it take for a package to fall from a hot-air balloon?

AI Thread Summary
A package dropped from a hot-air balloon ascending at 13 m/s from a height of 90 m takes approximately 4.25 seconds to reach the ground. The initial velocity of the package is 13 m/s, as it retains the balloon's speed at the moment of release. The final speed upon impact is calculated to be around 42.18 m/s, which is considered negative due to the downward direction. The discussion emphasizes using kinematic equations to solve for time and final velocity. Overall, the calculations confirm the correct application of physics principles in determining the package's fall.
suxatphysix
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Homework Statement


A hot-air balloon is ascending at the rate of 13 m/s and is 90 m above the ground when a package is dropped over the side.

(a) How long does the package take to reach the ground?


(b) With what speed does it hit the ground?



Homework Equations






The Attempt at a Solution



Tried 2a\Deltay=Vf^{2}-vi^{2}

I don't know if the initial velocity is 0 or 13m/s
 
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If the balloon is ascending at 13m/s, then vi=13m/s (with the sign convention that upwards is positive). Try using the equation y=vi*t+0.5a*t^2.
 
thanks, got part a right.

for b, I'm getting 42.18. is that right and would it be negative if it is going down?
used vf= vi + a*t and vf^2=vi^2+2*a*d
 
suxatphysix said:
thanks, got part a right.

for b, I'm getting 42.18. is that right and would it be negative if it is going down?
used vf= vi + a*t and vf^2=vi^2+2*a*d

Yes and yes; although you only need to use the second equation.
 
ty got it right
 
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