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Question:
I am teaching myself basic physics and have a question: If a helicopter is ascending vertically with a speed of 5.5 m/s. When it reaches a height of 105m a package is dropped from a window of the copter. How long does it take for the package to hit the ground.

I get an answer of 4.6 s, but I have the answers to these problems and it says it takes 5.22 s.

I figured the package dropped in the y direction is under the influence of gravity at 9.8 m/s2. It has to travel 105m to the ground. I figured: y=1/2a(t)2 = 4.6s Should I be thinking that the package has some initial velocity because the helicopter is ascending at 5.5 m/s? The problem doesn't state that the package is being thrown down by some initial velocity.



Replies:
You are correct in assuming that you have to take the initial velocity of the helicopter into account. Since the package was in the helicopter, it must also have been moving up at a velocity of 5.5 m/s. It doesn't really matter whether it was a constant velocity or not, just that this is the velocity of the object at the time it was released and the forces of the helicopter stopped acting on it.

There are probably several ways of solving this problem, but here is how I went about it:

1. Set up the y axix so positive is up, down is negative, so gravity is -9.8m/s2

2. Figure out the time it takes the object to stop rising:

v = vi + at (vi = initial velocity)

0 = 5.5 - 9.8m/s2 t

-5.5/-9.8 s = t

t = .56s

3. Figure the distance it traveled up in this time

d = 1/2(vi + v)t
d = 1/2(5.5m/s)(.56s)
d = 1.54m

4. Add the distance it traveled up to get the total distance it has to fall
106.54m

5. Figure the time based on the new distance, (distance is negative):

d = vi t + 1/2 a t2

-106.54 = 0 t + 1/2 -9.8m/s2 t2

-106.54 = -4.9m t2

t = sqrt(21.74)
t = 4.66 s

6. Add the time it took to stop rising to the time it took to fall

4.66s + .56s = 5.22s

Hope this clarifies the problem.

Thanks,

--ET


Yes, you have to take into account the upward velocity that the package has because it is associated with the copter. There will be a time and a distance associated with it going up at 5.5 m/sec until it reaches 0 m/sec velocity before it falls back toward earth. Notice I said a time and a distance. I imagine that is where your .6 sec difference lies. Another way to think of it is if you throw a baseball straight up in the air. It has a time associated with it rising above you, a point where it reaches 0 velocity, and then a return to the ground. The only difference is it is the helicopter creating the velocity and not your arm. Good luck, and feel free to visit us again if you have more questions. Physics is a fun subject to learn in my opinion.


The complete equation for linear motion is:
x = x(initial) + v(initial)t + (1/2)at^2

You are starting with an initial velocity of +5.5 m/s. Your acceleration is -9.8 m/s^2

Your stated equation does not take into account the initial velocity of the object.

---Nathan A. Unterman


Yes. The initial velocity of the package is the same as the velocity of the helicopter. The package is "dropped" from the window, which means that it is let go and allowed to fall. Relative to the ground, the package will first rise and then fall to the ground.

The velocity of the package at time t relative to the ground will be

v(t) = 5.5 m/s -9.8 ms^(-2) t

The height of the package above the ground at time t will then be

y(t) = 5.5 ms^(-1)t - 0.5 [9.8 ms^(-2)] t^2 + 105 m

and you need to solve this quadratic equation for when y = 0.

You might want first to verify that y = 0 (within roundoff error) when t = 5.22 s.

Richard Barrans Jr., Ph.D.


Hello, Since the package is on board, it is traveling at the same velocity as the helicopter. If you are in the helicopter and ascending, it "appears" to you that the initial velocity of the zero. Indeed, it initial relative velocity is zero. If you are on the ground, on the other hand, you will see the package ascending after the "drop", coming to standstill, and then dropping. So use the question: y =(1/2)gt^2+v, where v is the initial velocity.

AK Dr. Ali Khounsary



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