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?

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.

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.