Vfx: 12m/sVfy: -9.81m/sRubber Duckie Drop Homework

  • Thread starter Thread starter icurryx3
  • Start date Start date
  • Tags Tags
    Drop Rubber
AI Thread Summary
A hot air balloon ascends at 8 m/s when a rubber duckie is dropped, falling for seven seconds before bouncing off a roof and hitting the ground. The key equations for solving the problem involve vertical motion, specifically y(t) = y_0 + Vy_0t + 1/2at^2. The student initially struggled with setting up the equations but eventually resolved the calculations for the distances and velocities involved. The discussion highlights the importance of understanding the motion equations in physics problems. Ultimately, the student successfully solved all parts of the homework assignment.
icurryx3
Messages
19
Reaction score
1

Homework Statement


A hot air balloon containing an AP Physics student ascends vertically at a constant speed of 8 m/s. While ascending, he accidentally drops his rubber duckie from the gondola of the ballong. Seven seconds after it is dropped the duckie bounces off a roof at point A and hits the ground at point B as shown in the sketch at the right.

A. Find the distance from the point where the duckie is dropped to point A
B. Find the speed of the sweet little rubber duckie just before it hits the roof

Suppose that immediately after bouncing off the roof, the velocity of the rubber duckie is 12m/s 37 degrees BELOW the horizontal and that point B is horizontal distance of 24 meters from point A.

C. Find how far point A is above point B
D. Find the duckie's time of flight between points A and B
E. Find the Velocity of the duckie the instant before it strikes the ground at Point B

Homework Equations


V = Vo+at
V^2 = Vo^2 + 2ad
d = VoT + 1/2at^2

The Attempt at a Solution


Currently just filled out my DxVxT and VoyVyadyt to the following:
Voy: 8m/s
Vy: ?
a: -9.81m/s^2
dy:?
T:?
 
Physics news on Phys.org
icurryx3 said:

Homework Statement


A hot air balloon containing an AP Physics student ascends vertically at a constant speed of 8 m/s. While ascending, he accidentally drops his rubber duckie from the gondola of the ballong. Seven seconds after it is dropped the duckie bounces off a roof at point A and hits the ground at point B as shown in the sketch at the right.

A. Find the distance from the point where the duckie is dropped to point A
B. Find the speed of the sweet little rubber duckie just before it hits the roof

Suppose that immediately after bouncing off the roof, the velocity of the rubber duckie is 12m/s 37 degrees BELOW the horizontal and that point B is horizontal distance of 24 meters from point A.

C. Find how far point A is above point B
D. Find the duckie's time of flight between points A and B
E. Find the Velocity of the duckie the instant before it strikes the ground at Point B

Homework Equations


V = Vo+at
V^2 = Vo^2 + 2ad
d = VoT + 1/2at^2

The Attempt at a Solution


Currently just filled out my DxVxT and VoyVyadyt to the following:
Voy: 8m/s
Vy: ?
a: -9.81m/s^2
dy:?
T:?

You haven't really shown much of an attempt at the solution yet. What is the equation for the vertical position of the ducky as a function of time?
 
Would that not be the VoyVyaDyT?
Or maybe it is DxVxT. I'm really unsure right now; need some info and I can solve the rest.
 
icurryx3 said:
Would that not be the VoyVyaDyT?
Or maybe it is DxVxT. I'm really unsure right now; need some info and I can solve the rest.

Sorry, I have no idea what any of that means.

You have the beginning of the right equation when you wrote "d = VoT + 1/2at^2", but it is incomplete. What is the full equation for y(t) in terms of y_0, Vy_0, acceleration and time...?
 
Ah so would it be "d = (8.0)t + 1/2(-9.81)(t)^2"
 
icurryx3 said:
Ah so would it be "d = (8.0)t + 1/2(-9.81)(t)^2"

Closer, but still some issues.

First, I would use y(t) on the LHS of the equation instead of d. y(t) is much more descriptive of what is going on (the ducky is moving vertically with time).

Second, you have left off the initial y position of the ducky, y(0). Where does that go in the equation?

Once you get those fixed, you have an equation for the vertical position y(t) of the ducky with respect to time. What else are you told in the problem statement? How can you use those facts and this equation to come up with the first answer to the problem? :smile:
 
Hmm..
So the initial position would be 0 so the equation would come out to be:
0 = (8.0)t + 1/2(-9.81)t^2
Correct?
 
icurryx3 said:
Hmm..
So the initial position would be 0 so the equation would come out to be:
0 = (8.0)t + 1/2(-9.81)t^2
Correct?

No. Can you explain why you wrote that equation in response to my hints?
 
So what I initially planned to do first was to find the distance of the balloon itself, leaving the bigger picture apart. I thought that if I found how high the balloon is, I could find the duck falling and speed easily.
This was my plan:
Balloon: Vo = 8.0 m/s
Acceleration = -9.81m/s^2
However, I need one more factor to begin solving for any variable. I thought that since it took the duck 7 seconds to fall, I could just make the time of the balloon as 7 seconds.
Where did I go wrong?
 
  • #10
Sorry, I'm not sure what some of the terms you state actually mean.
 
  • #11
This is the equation you need to write and fill out. It is commonly used in problems like this one:

y(t) = y_0 + {Vy_0}t + \frac{1}{2}{a_y}t^2
 
  • #12
Nevermind, I have solved all of the problems. The equations were quite easy to be honest, just couldn't make out a mental picture.
Thanks for the help!
 
  • Like
Likes berkeman
Back
Top