A helicopter is rising at 5 m/s when a bag is dropped. After 2 seconds what is the bag's velocity? what is the bag's position? and how far below the helicopter is the bag?
Well, well, well! Seems like the bag decided to take a little leap of faith, huh? Let's break it down, shall we?
After 2 seconds, the bag's velocity will be the same as the helicopter's because it has dropped straight down. So, the velocity of the bag at that point will also be 5 m/s.
Now, in terms of position, the bag would have traveled a distance of 10 meters in those 2 seconds, assuming it fell without any air resistance trying to slow it down. That's because the velocity of the bag is constant, so it would have fallen 5 meters in the first second and another 5 meters in the second second.
As for how far below the helicopter the bag is, let's see. Since the bag dropped for 2 seconds at a constant velocity of 5 m/s, it would be 10 meters below the helicopter. So, look down, my friend, and you'll find that bag down there, 10 meters below the helicopter!
To find the bag's velocity after 2 seconds, we need to consider the acceleration due to gravity (typically denoted as -9.8 m/s²) acting on the bag. The bag is being dropped, so it only experiences vertical motion and its initial velocity is 0 m/s.
First, let's find the bag's velocity after 2 seconds. We know that the bag is only moving vertically and gravity is pulling it downward, so we can use the following equation:
v = u + at
Where:
v = final velocity
u = initial velocity (0 m/s in this case)
a = acceleration due to gravity (-9.8 m/s²)
t = time (2 seconds in this case)
Plugging in the values, we get:
v = 0 + (-9.8) * 2
v = -19.6 m/s
The bag's velocity after 2 seconds is -19.6 m/s, where the negative sign indicates that it is moving downward.
Now, let's find the bag's position after 2 seconds. Since the bag's initial position is not given, we can assume it starts at a height of 0 meters (relative to the helicopter).
To calculate the position, we use the equation:
s = ut + (1/2)at²
Where:
s = position
u = initial velocity (0 m/s)
t = time (2 seconds)
a = acceleration due to gravity (-9.8 m/s²)
Plugging in the values, we get:
s = 0 * 2 + (1/2) * (-9.8) * (2^2)
s = 0 + (1/2) * (-9.8) * 4
s = 0 + (-4.9) * 4
s = -19.6 meters
The bag's position after 2 seconds is -19.6 meters. Note that the negative sign indicates that it is below the starting point (which was taken as 0 meters).
Lastly, to find how far below the helicopter the bag is after 2 seconds, we consider the bag's position. Since the starting point is taken as 0 meters, we can conclude that the bag is 19.6 meters below the helicopter at this time.
In summary:
- The bag's velocity after 2 seconds is -19.6 m/s (moving downward).
- The bag's position after 2 seconds is -19.6 meters (below the starting point).
- The bag is 19.6 meters below the helicopter after 2 seconds.