A hot air balloon is traveling vertically upward at a constant speed of 3.9 m/s. When
it is 10 m above the ground, a package is
released from the balloon.
After it is released, for how long is the
package in the air? The acceleration of gravity
is 9.8 m/s
2
.
Answer in units of
Y = 10 + 3.9 t - (g/2) t^2 = 0
Subsitute in the g value and solve the equation. Take the positive root. (There will also be a negatve solution)
To find the time the package is in the air, we first need to find the time it takes for the package to fall to the ground.
We can use the equation of motion for vertical motion:
s = ut + (1/2)at^2
Where:
s = vertical displacement (change in height)
u = initial velocity (0 m/s, as the package is released with no initial velocity)
a = acceleration (acceleration due to gravity, -9.8 m/s^2 in the downward direction)
t = time
In this case, the vertical displacement s is equal to the initial height of the package, which is 10 m, and we need to solve for t.
So, the equation becomes:
10 = 0 * t + (1/2) * (-9.8) * t^2
Simplifying the equation, we have:
10 = -4.9 * t^2
Divide both sides of the equation by -4.9:
-2.04 = t^2
Taking the square root of both sides:
t = ā(-2.04)
Since time cannot be negative in this context, we discard the negative solution:
t = ā2.04 ā 1.428 seconds
Therefore, the package will be in the air for approximately 1.428 seconds.