Time interval until a ball returns to its orign

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Discussion Overview

The discussion revolves around the problem of determining the time interval until a ball returns to its origin when thrown from a balloon that is rising at a constant speed. Participants explore the relationships between the velocities of the balloon and the ball, as well as the equations of motion involved in the scenario.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a solution but does not clarify the meaning of the symbols used in their equations.
  • Another participant questions the validity of a trigonometric calculation, suggesting that a mistake was made by treating degrees as radians.
  • Concerns are raised about the initial velocity definitions, with multiple interpretations proposed for the symbols ##v_0## and ##v_y##.
  • A hint is provided regarding the scenario of throwing the ball in a rising elevator, suggesting that the velocity of the elevator can be inferred from the time it takes for the ball to return.
  • A participant calculates the total distance traveled by the ball, incorporating both the balloon's upward velocity and the effects of gravity.

Areas of Agreement / Disagreement

Participants express disagreement regarding the calculations and interpretations of the symbols used. There is no consensus on the correct approach or solution to the problem.

Contextual Notes

Participants have not clearly defined all variables, leading to potential misunderstandings. The discussion includes unresolved mathematical steps and assumptions about the frame of reference.

tremain74
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Homework Statement
I have a problem that I am working on from my Engineering Dynamics Book and I want to see if I am on the right path. A person in a balloon rising with a constant velocity of 4 m/s propels a ball upward with velocity of 1.2 m/s relative to the balloon. After what time interval will the ball return to the balloon? Answer: t = 0.245 seconds.
Relevant Equations
I used the equation vy = -9.8*t + v0*sin0(sin of theta).
This is my solution below.

PXL_20241002_163129761.jpg
 
Last edited by a moderator:
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Hi,

Do I really see a ##\sin(90)=0.89## in your picture?

(you should typeset the math in your post using ##\LaTeX ##!)

Then: The balloon goes up with 4 m/s, and the ball is thrown up with 1.2 m/s relative to the balloon.
So how can you write ##v_0=4## m/s ?

What equation are you solving (or rather: not solving, since you get a negative t) ?

##\ ##
 
Last edited:
@tremain74, in addition to what @BvU has already said...

##\sin (90 \text { radians}) = 0.894## approx. You have forgotten to switch your calculator to degrees mode, so it is treating '90' as 90 radians. But in any case, you need to know that ##\sin (90^0) = 1## without using a calculator!

I can’t follow the logic of your working. In addition you have not made clear what your symbols mean, for example:

Is ##v_0##:
- the initial velocity of the ball relative to the balloon?
- the initial velocity of the ball relative to the ground?
- the velocity of the balloon?
- something else?

Is ##v_y##:
- the velocity of the ball relative to the balloon after some time, ##t##?
- the velocity of the ball relative to the ground after some time, ##t##?
- the initial velocity of the balloon relative to the ground?
- something else?

If you deliberately change the meaning of a symbol mid-working (you shouldn't!) you need to state this.

Rethink and try again. Here's a big hint:

You are in an elevator rising at a constant velocity. You throw a ball up at a speed of 1.2m/s relative to you and measure the time it takes to come back to you. Can you find the elevator's velocity from your result?
 
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The balloon up speed is 4 m/s constant. Then the space up to
the point of return of ball it is 4*t.
The total space of the ball is (4+1.2)*t-9.8*t^2/2
 

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