During liftoff, a hot-air balloon accelerates upward at a rate of 3.3m/s^2. The balloonist drops an object over the side of the gondola when the speed is 15m/s.
Part A: What is the magnitude of the object’s acceleration after it is released (relative to the ground)?
Part B: How long does it take to hit the ground?
F = m a
-m g is the only force on the object
-m g = m a
a = -g = -9.81 m/s^2
now where is the balloon?
v = 0 + 3.3 t
15 = 3.3 t
t = 4.55 seconds to reach v = 15
h = (1/2)a t^2 = (1/2)(3.3)(4.55^2) = 34.2 m high
so we release the rock at 34.2 meters while moving up at 15 m/s
a = -9.81
h = 34.2 + 15 t - 4.9 t^2
when is h = 0 ? (hits ground)
4.9 t^2 -15 t - 34.2 = 0
t = [ 15 +/ - sqrt (225 + 670) ] /9.8
t = [ 15 + 29.9 ]/9.8
t = 4.58 seconds
Part A: The magnitude of the object's acceleration after it is released is equal to the acceleration due to gravity, which is approximately 9.8 m/s^2. This is because once the object is released, it is only under the influence of gravity, and there is no other force acting on it.
Part B: To find the time it takes for the object to hit the ground, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance or height travelled by the object (in this case, the height of the balloon)
u = initial velocity of the object (15 m/s)
a = acceleration of the object due to gravity (-9.8 m/s^2, taking downward direction as negative)
t = time taken
Since the object is dropped, its initial velocity (u) is zero, and the equation simplifies to:
s = (1/2)at^2
Assuming the initial height of the balloon is also zero, we can rearrange the equation to solve for time:
t = sqrt(2s/a)
Given the initial velocity of the balloon (15 m/s), we need to determine how high the balloon has ascended before the object is dropped.
Using the equation of motion for the balloon's upward motion:
v^2 = u^2 + 2as
Where:
v = final velocity of the balloon (when the object is dropped)
u = initial velocity of the balloon (0 m/s)
a = acceleration of the balloon (3.3 m/s^2, taking upward direction as positive)
s = height gained by the balloon before the object is dropped
Rearranging the equation and solving for s:
s = (v^2 - u^2) / (2a) = (15^2 - 0^2) / (2*3.3) = 225 / 6.6 = 34.09 meters (rounded to 2 decimal places)
Now we can substitute the values into the equation for time:
t = sqrt(2s/a) = sqrt(2*34.09 / 9.8) = sqrt(69.1847) ≈ 8.31 seconds (rounded to 2 decimal places)
Therefore, it takes approximately 8.31 seconds for the object to hit the ground after it is released.
To solve Part A, we need to find the magnitude of the object's acceleration after it is dropped. Since the hot-air balloon is accelerating upward, the object will still have this upward acceleration relative to the balloon. Therefore, the object's acceleration relative to the ground will be the sum of the balloon's acceleration and the acceleration due to gravity.
The balloon's acceleration is given as 3.3 m/s^2 upward. The acceleration due to gravity is approximately 9.8 m/s^2 downward. Since these accelerations act in opposite directions, we need to subtract the acceleration due to gravity from the balloon's acceleration.
Magnitude of the object's acceleration = magnitude of balloon's acceleration - magnitude of acceleration due to gravity
= 3.3 m/s^2 - 9.8 m/s^2
= -6.5 m/s^2
Therefore, the magnitude of the object's acceleration after it is released (relative to the ground) is 6.5 m/s^2 downward.
To solve Part B, we can use the equations of motion. The object is being dropped from rest at a height of the gondola. We need to find the time it takes for the object to reach the ground.
We can use the following equation:
s = ut + (1/2)at^2
Where:
s = distance or height (in this case, the distance from the gondola to the ground)
u = initial velocity (which is 0 m/s because the object is being dropped from rest)
a = acceleration (in this case, -6.5 m/s^2 due to gravity)
t = time
Since the initial velocity is 0, the equation simplifies to:
s = (1/2)at^2
The distance s is the height of the gondola at which the object is dropped. We can substitute the given values into the equation.
s = 15 m
a = -9.8 m/s^2
15 = (1/2)(-9.8)t^2
Simplifying the equation further:
15 = -4.9t^2
Divide both sides by -4.9:
t^2 = -15 / -4.9
t^2 ≈ 3.061
Taking the square root of both sides:
t ≈ √3.061
t ≈ 1.748 seconds (rounded to three decimal places)
Therefore, it takes approximately 1.748 seconds for the object to hit the ground after being dropped from the gondola.