Velocity in relation to position

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Homework Help Overview

The problem involves a hot-air balloon rising at a constant velocity and a camera being tossed upward from the ground. The objective is to determine the minimum initial speed required for the camera to reach the passenger in the balloon, who is at a specific height above the point of toss.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to find the initial speed of the camera. There is a focus on the relationship between the velocities and heights of the balloon and the camera. Some participants express confusion regarding the correct interpretation of the balloon's constant rate of ascent and the appropriate application of the kinematic equations.

Discussion Status

There is ongoing exploration of the problem with participants questioning the assumptions made about the balloon's velocity and the units used in the calculations. Suggestions have been made to express the altitude of both the balloon and the camera as functions of time to facilitate solving for the initial velocity.

Contextual Notes

Participants note the ambiguity in the description of the balloon's motion, specifically whether it is constant velocity or constant acceleration. There is also mention of the need for clarity in units for the given data.

destinc
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A hot-air balloon has just lifted off and is rising at the constant rate of 1.8 . Suddenly one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward. If the passenger is 3.1 above her friend when the camera is tossed, what is the minimum initial speed of the camera if it is to just reach the passenger? (Hint: When the camera is thrown with its minimum speed, its speed on reaching the passenger is the same as the speed of the passenger.)


I tried to use V^2=Vi^2 + 2aΔX to solve for Vi
I plugged in (1.8)^2=Vi^2 +2(-9.8)3.1
and got Vi=8m/s but that answer is wrong. Where did I go wrong?
 
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destinc said:
A hot-air balloon has just lifted off and is rising at the constant rate of 1.8 . Suddenly one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward. If the passenger is 3.1 above her friend when the camera is tossed, what is the minimum initial speed of the camera if it is to just reach the passenger? (Hint: When the camera is thrown with its minimum speed, its speed on reaching the passenger is the same as the speed of the passenger.)

I tried to use V^2=Vi^2 + 2aΔX to solve for Vi
I plugged in (1.8)^2=Vi^2 +2(-9.8)3.1
and got Vi=8m/s but that answer is wrong. Where did I go wrong?
First of all, you need some units on the given data.

Then, "rising at the constant rate of 1.8" could mean a constant velocity, or a constant acceleration. Again, units would help clear up this ambiguity.
 
sorry, balloon is rising at constant rate of 1.8m/s.
balloon height is 3.1m.
for the formula I used (1.8m/s)^2=Vi + 2(-9.8m/s^2)3.1m
 
destinc said:
sorry, balloon is rising at constant rate of 1.8m/s.
balloon height is 3.1m.
for the formula I used (1.8m/s)^2=Vi + 2(-9.8m/s^2)3.1m
You are mixing the velocity and altitude of the balloon with the acceleration of the camera. Also vi should be squared in this formula. Beside that, I doubt that this kinematic equation is the most helpful in solving this problem.

Express the altitude of the balloon as a function of time, t, where t = 0 seconds at the moment at which the balloon's altitude was 3.1 m, which is also the moment the camera was tossed up .

Express the altitude of the camera as a function of time, t .

Equate the two & solve for t. The result will depend on the initial velocity of the camera.
 

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