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A hot air balloon is rising upward with a constant speed of 2.50m/s. When the balloon is 3.0 m above the ground, the balloonist drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?
PS. Do you mind explaining why you chose a certain formula or why you plugged in certain values? 

Sagot :

samuelhoang
Let's assume that the balloonist dropped the compass over the the side as the balloon still rising upward at 2.5m/s

So you have:
Vi = -2.5 m/s (because the compass is still going with the balloon as it starts to fall down)
d = 3 m
a = 9.81 m/s^2

At first you use this formula to find the Vf (finally velocity):
Vf^2 = Vi^2 + 2ad
Vf^2 = 6.25 + 2(9.81)(3)
Vf^2 = 65.11
Vf = 8.07 m/s

Finally you use this formula to find the time:
Vf = Vi +at
8.07 = -2.5 + (9.81)t
10.57 = (9.81)t
t = 1.08 s

Your final answer is 1.07 seconds.

AL2006
In the first answer submitted, Samuel got a good answer.  But for the life of me,
I'm having trouble following his solution.  I'm not that great at math and physics,
so here's how I did it :

I always call 'up' positive and 'down' negative, and I use one single
formula for the height of anything that's tossed or dropped:

H = Height at any time
H₀ = height when it's released  =  3m
V₀ = vertical speed when it's released  =  2.5 m/s
T = time after it's released
G = acceleration of gravity  =  -9.8 m/s²

H = H₀ + V₀T - 1/2 G T²

H = 3 + 2.5T - 4.9T²

That's the height of the compass at any time after it's dropped, and
we simply want to know the time ' T ' when  H = 0 (it hits the ground).

- 4.9T² + 2.5 T + 3  =  0

That's a perfectly good quadratic equation, which you can solve for ' T ' .
The solutions are    T = -0.567sec  and  T = 1.078sec.

The one with physical significance in the real situation is  T = 1.078sec.

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