Need help with calculating Gravitational Constant

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Homework Help Overview

The discussion revolves around calculating the gravitational constant on a foreign planet based on the fall of a bomb from a height of 200 meters. The bomb's last 100 meters of descent took 2.5 seconds, prompting questions about the acceleration due to gravity in this context.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to create equations for the total fall and the last segment of the fall but encounters difficulties due to multiple variables. Some participants suggest using kinematic equations to express the motion and isolate variables.

Discussion Status

Participants are exploring different equations and methods to isolate the acceleration. There is a mix of suggestions regarding how to manipulate the equations, but no consensus has been reached on a specific approach or solution.

Contextual Notes

There is a mention of the gravitational constant being universal, which some participants seem to question in the context of the problem. Additionally, the original poster expresses uncertainty about isolating the variable for acceleration.

JohnDoeJD
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Homework Statement



I would like some help with this question:

A bomb is dropped on a foreign planet. The bomb is dropped from a height of 200m. The bomb traveled the last 100m of the fall in 2.5 seconds. What is the acceleration from the gravitational constant on this planet?


2. The attempt at a solution

I tried making two separate equations, one for the total fall (200m) and one for the 100m fall and combining them but I couldn't get anything because I had too many variables.
 
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JohnDoeJD said:
What is the acceleration from the gravitational constant on this planet?

do you mean "and" ?
 
Yes, that's what I meant sorry.
 
write s = ut + .5 at^2 for (t - 2.5) and t seconds
t it the total time of fall

substract 2 eqn's and substitute st - st - 2.5 = --- ?
 
JohnDoeJD said:
Yes, that's what I meant sorry.

gravitational constant remains same everywhere

thats why it is called "UNIVERSAL" gravitational constant !

EDIT:

confirm it yourself by finding its value in this question
 
cupid.callin said:
write s = ut + .5 at^2 for (t - 2.5) and t seconds
t it the total time of fall

substract 2 eqn's and substitute st - st - 2.5 = --- ?

I did that originally but I couldn't figure out how to isolate A. This is what I had:

100 = 0(t) + 0.5at^2 - 0(t-2.5) + 0.5a(t-2.5)^2
 
rearrange this eqn to a simpler form

use another eqn: 200 = .5at^2

use this eqn to subs. t in first eqn
yu'll get something in √a and a

consider √a as y and and a as y^2 and solve this just like a quadratic eqn!
 
Thank you.
 

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