How Long Does It Take for a Released Mailbag to Hit the Ground?

[Saterial]

Homework Statement

The height of a helicopter above the ground is given by h=3.20t^3, where h is in meters and t is in seconds. After 2.05s, the helicopter releases a small mailbag. How long after it's release does the mailbag reach the ground?

Homework Equations

Δd=v1Δt + 1/2aΔt^2

The Attempt at a Solution

At first when I looked at this question, I thought it was really easy, however it didn't work out well.

Δd=3.20t^3
V1=0
a=9.8m/s^2

Δd=v1Δt+1/2aΔt^2
3.20t^3=-4.9t^2

My approach to this is most likely wrong but I can't figure out why. I picked an arbitrary direction of down as positive and my thoughts on this question are, can't I simply divide 3.20t^3 by 4.9t^2 to cancel out the t leaving you with just t? However, that didn't work. Additionally, I assumed that the 2.05s is not used, but if there is, I can't see how it would be of relevance to the question.

 

[Forty-Two]

You have the right idea about which equations to use.

Your definitions of your t's is not correct however. The t in the equation, h=3.20t^3, is being used to refer to the height of the helicopter at a certain time t.

What you are trying to solve for however is a different time, the time it takes the bag to fall to the ground. These two times are not necessarily equal.

Hope this helps and let me know if you need more guidance.

 

[Saterial]

Is there a purpose of the 2.05s ? I don't see how it's of any relevance to the time it takes for the mailbag to fall down.

 

[hogrampage]

Is there a purpose of the 2.05s ? I don't see how it's of any relevance to the time it takes for the mailbag to fall down.

Yes, that can be used to find the height at which the helicopter is when the bag is launched. This will be your y value (the height it must fall). Use that to solve for t using:

y = 1/2gt^2

(Note: The actual equation is y-y_o = v_osin(theta)t - 1/2gt^2, but it is a launch of 0 degrees, so sin(0) = 0. And, I just replaced -1/2gt^2 with 1/2gt^2 because the bag is falling downward - in the direction of gravity).

Someone correct me if I'm wrong.

 

[rwisz]

Your approach is close, but there are a few things that need to be corrected. First, the equation Δd=v1Δt+1/2aΔt^2 is for constant acceleration, which is not the case in this scenario. The helicopter's height is changing at a rate of 3.20t^3, which means the acceleration is also changing. Therefore, we cannot use this equation.

Instead, we can use the kinematic equation h= h0 + v0t + 1/2at^2, where h0 is the initial height, v0 is the initial velocity, and a is the acceleration. In this case, h0 is the height of the helicopter at t=0, which is 0 since it starts at the ground. v0 is also 0 since the helicopter is initially at rest.

So, the equation becomes h= 0 + 0(t) + 1/2a(t)^2, which simplifies to h= 1/2at^2.

Now, we can plug in the given values to solve for the time at which the mailbag reaches the ground. We know that h= 0 when the mailbag reaches the ground, so we can set the equation equal to 0 and solve for t:

0= 1/2(9.8)(t)^2
0= 4.9(t)^2
t= √(0/4.9)
t= 0 seconds

This means that the mailbag reaches the ground at the same time it is released, which is 2.05 seconds after the helicopter starts its descent. So, the answer to the question is 2.05 seconds.

In summary, the key to solving this problem is to use the correct kinematic equation and to remember that the acceleration is not constant in this scenario. Also, using the given information of 2.05 seconds can help to solve for the time at which the mailbag reaches the ground.