What is the gravitational acceleration constant of Oz?

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Homework Help Overview

The problem involves calculating the gravitational acceleration constant in a fictional context where a character drops an object from a height. The scenario includes specific measurements related to the object's fall, including its mass, the height of the drop, and the time taken for a portion of the fall.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using kinematic equations to relate distance, time, and acceleration. There are attempts to set up equations based on the given data, but some express difficulty in isolating variables or finding a complete solution. Questions arise about the implications of quadratic factors and the need for additional equations.

Discussion Status

The discussion is ongoing, with participants sharing their attempts to formulate equations and expressing uncertainty about the next steps. Some guidance has been offered regarding the need for multiple equations to solve for the unknowns, but no consensus or resolution has been reached.

Contextual Notes

Participants are working within the constraints of a fictional scenario, which may influence their interpretations of the physics involved. The problem setup includes specific measurements that may not align with typical gravitational scenarios.

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Homework Statement



Lindsay was on her way to visit her grandmother in Kansas when suddenly a tornado picked her up and transported her to Oz. While standing beside a 198m cliff, she dropped a yellow brick over the side. It has a mass of 1.9Kg. As it was plummeting to the ground, she decided to time the last 50m of its fall. She measured 1.5s on her stop watch. She later impressed the wizard by correctly calculating the gravitational acceleration constant of Oz. What was it?

Homework Equations



d = V1t + 0.5(a)(t)^2
d = V2t - 0.5(a)(t)^2

d[total] = 198m
d[timed] = 50m
m[timed] = 1.9Kg
t = 1.5s


The Attempt at a Solution



I tried to use all the equations I know, but I'm always missing one variable to complete it. Could you guys help me out and give me a starting point so I can finish it? Greatly appreciated.

lewl'.JPG
 
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Fill in the blanks:

First postion: d = ?; time = t
Second position: d = 198 m; time = t + 1.5 s
 
Thanks Al, so far, I got this:

d=v1t+0.5(a)(t)^2
198m=(0.5)(a)(t+1.5)^2
198m=(0.5)(a)(t^2+3t+2.25)

Where do you go from there? It looks like I need another hint.

Factoring that quadratic gives me a double zero at -1.5, what does that mean?

I was thinking... 0=(0.5)(a)((2x+3)^2)-198m, but it doesn't make sense.
 
SPH4U said:
d=v1t+0.5(a)(t)^2
198m=(0.5)(a)(t+1.5)^2
198m=(0.5)(a)(t^2+3t+2.25)
That describes the second point. You also need an equation describing the first point.

Then you'll have two equations and two unknowns.
 

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