1-Dimensional Motion Helicopter Problem (PPB)

[Joshpho]

Homework Statement

From People's Physics Book, ch. 3:

A helicopter is traveling with a velocity of 12 m/s directly upward. Directly below the helicopter is a very large and very soft pillow. As it turns out, this is a good thing, because the helicopter is lifting a large man. When the man is 20 m above the pillow, he let's go of the rope.

a. What is the speed of the man just before he lands on the pillow?
b. How long is he in the air after he let's go?

c. What is the greatest height reached by the man above the ground? (Hint: this should be greater than 20 m. Why?)

d. What is the distance between the helicopter and the man three seconds after he let's go of the rope?

Homework Equations

For a. I got 23 m/s which is correct and b, 3.6 s which is correct as well. For c I chose the equation

x(t)=x0+v0t+1/2at^2

The Attempt at a Solution

And plugged things in like so... (where upwards is positive)
x(t)=20m+12m/s*t+1/2(9.8m/s^2 )(t^2 )

And then I just started in plugging in times. 1 second was the highest x(t) value I could find, at 27.1 meters, but the book says the correct answer (max height) is 28 m. Am I missing something?

Help with d would also be helpful.

Additional question: Is there a resource for the people's physics book that has explanations for problems?

 

[CFede]

Pugging random times probably won't get you to an exact answer.

The moment when the man is at the greatest height is when his velocity reaches 0. After he let's go of the rope, he keeps going up, because of the initial speed of 12m/s, after a time t=12/9.8=1.22s he stops going up ans starts to fall, this is the moment of greatest height and is the time you should put into the equation to get the answer.

I'm quite sure this is what you should do, however, i don't get the 28m the book says.

Actually, if you look at it more carefully and try to find a time for which the height is 28m, there won't be a solution so, either the equation is not well written o the book is wrong. My guess is the book is wrong.

As for the letter d all you should do is to put t=3s in the equation you wrote. I think the answer is 8.1m.

 

[cepheid]

Welcome to PF,

For part c: It's as though he starts at an initial height of 20 m, with an initial upwards velocity of v₀ = 12 m/s. Now, when he reaches his max height, his vertical velocity will have been reduced to 0 m/s. Think about it: the maximum height must be the point at which he momentarily comes to a stop. If he didn't come to a stop, then he would still be moving upwards, and so that wouldn't be his maximum height!

So, one approach that you can use here is to use the kinematics equation that tells you the velocity versus time:

Δv = at

v - v₀ = at = -gt

​

Set v = 0, solve for t, and then plug this into your expression for the distance vs. time to find the height at that instant.

However, a much easier method is to use conservation of energy, if you know about that.

EDIT: And I agree that 28 m is wrong.

 

[azizlwl]

For d.
Find displacement of the man for 3secs after detachment.
Take point of detachment as origin. Upward positive.
Find displacement of the helicopter too.
Add the magnitude of these displacements.

 

[Nickie]

First of all, great job on solving parts a and b correctly! For part c, you are on the right track with using the equation x(t)=x0+v0t+1/2at^2. However, the key to finding the maximum height reached by the man is to remember that at the highest point, the velocity of the man will be zero. This means that v0=0 in the equation. Also, the initial position (x0) should be the height at which the man is released from the rope, which is 20 m in this case. So the equation should look like this:

x(t)=20m+0*t+1/2(9.8m/s^2)(t^2)

Now, when you solve for x(t), you should get the maximum height of 28 m, as stated in the book.

For part d, you can use the same equation, but this time the initial position (x0) will be the distance between the helicopter and the man when he releases the rope. This distance can be found using the equation x(t)=x0+v0t+1/2at^2, where x(t) is the maximum height reached by the man (which we already found in part c), and v0 is the initial velocity of the man (which you can find using the equation v(t)=v0+at). Once you have the initial position, you can plug it into the equation x(t)=x0+v0t+1/2at^2, with t=3 seconds, to find the distance between the helicopter and the man.

As for your additional question, unfortunately, I am not aware of any specific resource for the People's Physics Book that provides explanations for the problems. However, there are many online forums and communities where you can ask for help and clarification on specific problems. You can also try reaching out to your teacher or peers for assistance. Practice and persistence are key in mastering physics problems!