During liftoff, a hot-air balloon accelerates upward at a rate of 2.5 . The balloonist drops an object over the side of the gondola when the speed is 12 .
a)What is the magnitude of the object’s acceleration after it is released (relative to the ground)?
Part B
What is the direction of the object’s acceleration after it is released (relative to the ground)?
What is the direction of the object’s acceleration after it is released (relative to the ground)?
downward
upward
horizontal
Part C
How long does it take to hit the ground?
Express your answer using two significant figures.
To find the answers to these questions, we need to analyze the motion of the object dropped from the hot-air balloon.
a) The magnitude of the object's acceleration after it is released (relative to the ground):
The object experiences two accelerations: the upward acceleration due to the hot-air balloon (2.5 m/s^2) and the downward acceleration due to gravity (9.8 m/s^2). Since the object is released, it is no longer affected by the upward acceleration from the hot-air balloon. Therefore, the magnitude of the object's acceleration after it is released is equal to the acceleration due to gravity (9.8 m/s^2).
b) The direction of the object's acceleration after it is released (relative to the ground):
Since the object is subject only to the acceleration due to gravity (9.8 m/s^2) after it is released, the direction of its acceleration will be downward. Therefore, the answer is downward.
c) To determine how long it takes for the object to hit the ground, we can use the kinematic equation:
s = ut + (1/2)at^2
where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the initial velocity is 12 m/s (the speed of the hot-air balloon when the object is dropped), the acceleration is the acceleration due to gravity (-9.8 m/s^2), and the distance traveled is unknown.
Assuming the object is dropped from rest (zero initial velocity), the equation becomes:
s = (1/2)at^2
Plugging in the values, we have:
0 = (1/2)(-9.8)t^2
Simplifying the equation, we get:
0 = -4.9t^2
This is a quadratic equation, and we are interested in the positive value of t since time cannot be negative. Solving for t, we find:
t = √(0/4.9) = 0 seconds
The resulting time is zero, indicating that the object hits the ground instantaneously after it is released.
a) The magnitude of the object's acceleration after it is released (relative to the ground) is 9.8 m/s^2. This is the acceleration due to gravity, which remains constant throughout the motion of the object.
b) The direction of the object's acceleration after it is released (relative to the ground) is downward. This is because gravity pulls objects towards the center of the Earth.
c) To calculate 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 (height from which the object is dropped)
u = initial velocity (0 m/s)
a = acceleration (9.8 m/s^2)
t = time
Since the initial velocity is 0 m/s, the equation simplifies to:
s = (1/2)at^2
Rearranging the equation to solve for time, we get:
t = √(2s/a)
Substituting the values given:
s = 12 m (height from which the object is dropped)
a = 9.8 m/s^2 (acceleration due to gravity)
t = √(2(12) / 9.8)
t ≈ 1.1 seconds
Therefore, it takes approximately 1.1 seconds for the object to hit the ground.