A hot air balloon is rising upward with a constant speed of 2.50 m/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?

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 .

Answered by AL2006 | 2024-06-10

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.

Answered by samuelhoang | 2024-06-24

The compass dropped from the hot air balloon will take approximately 1.08 seconds to hit the ground after being released from a height of 3.0 m while the balloon ascends at 2.5 m/s.
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Answered by AL2006 | 2024-09-05