A hot air balloon is traveling vertically upward
at a constant speed of 2.7 m/s. When
it is 11 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 s
Well, being a package, I'm sure it had some "air"borne dreams, but unfortunately, it was just a falling star. So, let's do some calculations to figure out how long it stayed in the air.
First, we know that the initial velocity of the package is the same as the hot air balloon's velocity, which is 2.7 m/s. Since it's being released, we can assume there is no more upward force acting on it.
Next, we need to determine the time taken for the package to fall to the ground. We can use the equation:
h = ut + (1/2)gt^2
where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.
We want to find t, so let's rearrange the equation:
(t^2)(1/2)g + (ut - h) = 0
Now, substitute the given values:
(1/2)(9.8)(t^2) + (2.7)(t) - 11 = 0
Solving this quadratic equation will give you the time t. Then, you can rejoice as you'll know how long the package took its brief flight!
To find the time for which the package is in the air, we can use the equations of motion.
First, let's find the time it takes for the package to reach the ground after it is released. We know that the vertical distance traveled by an object under constant acceleration is given by the equation:
s = ut + (1/2)at^2
Where:
s = vertical distance traveled (11 m)
u = initial vertical velocity (0 m/s, as the package is released with zero initial velocity)
a = acceleration due to gravity (-9.8 m/s^2, as it acts downward)
t = time
Rearranging the equation, we get:
t^2 = (2s) / a
Substituting the values, we have:
t^2 = (2 * 11) / 9.8
t^2 = 2.2449
Taking the square root of both sides, we find:
t ≈ 1.4982 s
So, it takes approximately 1.4982 seconds for the package to reach the ground. Since the package was released when the balloon was 11 m above the ground, the total time the package is in the air is 1.4982 seconds.
To find the time for which the package is in the air, we can use the following steps:
1. Calculate the time it takes for the package to reach the ground when dropped from a height of 11 m above the ground.
To do this, we can use the equation of motion:
Δy = V₀t + (1/2)gt²
where:
- Δy is the change in height (from 11 m to 0 m),
- V₀ is the initial velocity (which is 0 m/s as the package is dropped),
- g is the acceleration due to gravity (9.8 m/s²),
- t is the time we need to find.
Substituting the known values into the equation, we get:
-11 m = 0(t) + (1/2)(9.8 m/s²)t²
Simplifying the equation, we have:
-5t² = -11
Solving for t, we get:
t² = 11/5
t ≈ ± (√(11/5))
Since time can't be negative, we take the positive value:
t ≈ √((11/5))
2. Calculate the value of t.
Using a calculator, we can find the estimated value of t:
t ≈ √(2.2) ≈ 1.4832 s
Therefore, the package is in the air for approximately 1.4832 seconds.
Vi = 2.7 m/s
h = 11 + 2.7 t = 4.9 t^2
0 = 11 + 2.7 t - 4.9 t^2 at ground
4.9 t^2 - 2.7 t - 11 = 0
t = [ 2.7 +/- sqt (2.7^2 +44*4.9) ]/9.8