2.2. A hot air balloonist, rising vertically with a constant velocity of magnitude 5.00m/s releases a sandbag at an instant when the balloon is 40.0m above the ground as shown. After it is released, the sandbag is in free fall.
1. calculate the position and velocity of the sandbag at 0.25s and 1.00s after its release.
a. h = ho - (Vo*t + 0.5g*t^2)
h = 40 - (-5*0.25 + 4.9*0.25^3) = 40.9 m
V = Vo + g*t = -5 + 9.8*0.25=-2.55 m/s.
b. h = 40 - (-5*1 + 4.9*1^2) = 40.1 m.
V = -5 + 9.8*1 = 4.8 m/s.
To calculate the position and velocity of the sandbag at different times, we can use the equations of motion for objects in free fall.
First, let's establish the downward direction as negative. This means that any values calculated as positive will be above the ground, and any values calculated as negative will be below the ground.
Given data:
Initial velocity of the hot air balloonist: 5.00 m/s upward
Initial height of the balloonist: 40.0 m
Acceleration due to gravity: 9.8 m/s^2
We need to find the position and velocity at two different times: 0.25 seconds and 1.00 second.
Step 1: Calculate the time it takes for the sandbag to reach the ground after being released.
Since the balloonist is already going up at a constant velocity, the sandbag will take the same time to reach the ground as if it was dropped from rest.
We can use the equation: h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity, and t is the time.
Rearrange the equation to solve for t: t = sqrt(2h/g).
Using the given initial height of 40.0 m, the time it takes for the sandbag to reach the ground can be calculated as:
t = sqrt(2 * 40.0 / 9.8) ā 2.03 seconds
Step 2: Calculate the position and velocity at 0.25 seconds after release.
Since the sandbag takes 2.03 seconds to reach the ground, at 0.25 seconds it is still in the air.
To calculate the position at 0.25 seconds, we can use the equation: h = h0 + v0t + (1/2)gt^2,
where h is the final height, h0 is the initial height, v0 is the initial velocity, g is the acceleration due to gravity, and t is the time.
Let's begin with the position:
h = 40.0 + (5.00)(0.25) + (1/2)(9.8)(0.25)^2.
Simplifying the equation will give us the position of the sandbag at 0.25 seconds.
Next, let's calculate the velocity at 0.25 seconds using the equation: vf = v0 + gt,
where vf is the final velocity, v0 is the initial velocity, g is the acceleration due to gravity, and t is the time.
Using the given initial velocity as 5.00 m/s upward, we can calculate the velocity of the sandbag at 0.25 seconds using the equation above.
Step 3: Calculate the position and velocity at 1.00 second after release.
Since the sandbag takes 2.03 seconds to reach the ground, at 1.00 second it is still in the air.
Using the equations mentioned previously, we can calculate the position and velocity at 1.00 second in a similar manner to Step 2.
At this point, we have all the necessary information to calculate the position and velocity at both 0.25 seconds and 1.00 second after the release of the sandbag.