A hot air ballon is ascending at 12m/s and at 80 m above the ground, a package is dropped over teh sdie. How long does it take the package to hit the ground?
The equation for this would be h=g/2(t²)+12t+80. This works out to:
h=-4.9t²+12t+80
When the package hits the ground, t=0, so:
0=-4.9t²+12t+80
4.9t²-12t-80=0
49t²-120t-800=0
Solve for t
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To determine how long it takes for the package to hit the ground, we need to calculate the time it takes for the package to cover the distance of 80 meters while considering the initial velocity of the hot air balloon.
We can use the equation of motion that relates distance, initial velocity, time, and acceleration:
s = ut + (1/2)at^2
Where:
- s is the distance traveled (80m in this case)
- u is the initial velocity of the package (12m/s, since it was dropped from the hot air balloon)
- t is the time taken
- a is the acceleration due to gravity (-9.8m/s^2, because it is directed downwards)
Since the package is only influenced by the acceleration due to gravity, the equation reduces to:
s = ut + (1/2)gt^2
Substituting the given values, we have:
80m = 0t + (1/2)(-9.8m/s^2)t^2
This equation is a quadratic equation, we can rearrange it to solve for t. Multiplying through by 2 to eliminate the fraction, we have:
160m = -9.8m/s^2 * t^2
Rearranging the equation further:
t^2 = -160m / -9.8m/s^2
t^2 = 16.33s^2
Now, we can take the square root of both sides to solve for t:
t = sqrt(16.33s^2)
t ≈ 4.04s
Therefore, it takes approximately 4.04 seconds for the package to hit the ground after being dropped from the hot air balloon.