During liftoff, a hot-air balloon accelerates upward at a rate of 3.4 m/s^2 . The balloonist drops an object over the side of the gondola when the speed is 18 m/s.
What is the magnitude of the object’s acceleration after it is released (relative to the ground)?
You need to know how much mass was dropped, compared to the mass of the balloon. The buoyancy force will be the same, but the acceleration will increase in proportion to the reduction of mass.
Knowing the speed of the balloon when the mass was dropped does not provide the needed information.
i found the answer to the question it is 9.8 m/s^2. but how long would it take to hit the ground?
You are right. I misread the problem. I thought they were asking for the acceleration rate of the balloon after the object was released. A 'duh' moment.
Once released, the dropped object accelerates downward at the acceleration of gravity, g = 9.8 m/s^2
The time after liftoff when the object was released was:
t = V/a = 18/3.4 = 5.3 seconds.
The height it had risen at that time was
(1/2)at^2 = 47.6 m
Now you can calculate the additional time t' it takes for the dropped object to hit the ground.
y = 47.6 + 18 t' - 4.9 t'^2 = 0
Solve that quadratic for t' and take the positive root.
thank you very much
To find the magnitude of the object's acceleration after it is released relative to the ground, we need to first determine its initial velocity and the acceleration due to gravity.
Given:
Acceleration of the hot-air balloon (a) = 3.4 m/s^2
Velocity of the hot-air balloon (v) = 18 m/s (just before the object is dropped)
We know that the object will continue to move with the same velocity horizontally once it is dropped. However, the object will also experience the acceleration due to gravity acting downward.
The acceleration due to gravity (g) on Earth is approximately 9.8 m/s^2, directed downward.
Since the object is dropped from rest relative to the balloon, its initial velocity is zero. Therefore, the horizontal and vertical motion of the object are independent.
To find the object's acceleration after it is released, we need to separate the vertical motion from the horizontal motion.
The object will continue moving horizontally at a constant velocity, so its horizontal acceleration is zero.
The object will experience only the acceleration due to gravity in the downward direction, which gives its vertical acceleration. The magnitude of the object's acceleration after it is released (relative to the ground) is equal to the magnitude of the acceleration due to gravity, which is approximately 9.8 m/s^2.
Therefore, the magnitude of the object's acceleration after it is released (relative to the ground) is 9.8 m/s^2.