Answer :
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.
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.
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.
