Simple (was thought to be) Gravity Question

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Homework Help Overview

The problem involves a ball rolling up a slope, reaching a maximum displacement, and then rolling back down. The original poster is trying to determine the initial velocity of the ball based on given displacement and time.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the formula d=vt but questions the validity of their calculation when comparing it to the provided answer key. Other participants suggest using a different equation that accounts for acceleration, leading to discussions about the correct form of the equation and the implications of the word "constant" in the problem statement.

Discussion Status

Participants are actively engaging with the problem, questioning the assumptions made by the original poster and clarifying the equations involved. There is a recognition of the need to consider acceleration in the context of the problem, and some participants offer guidance on how to approach the situation without reaching a consensus on the solution.

Contextual Notes

The original poster mentions being new to physics and expresses concern about the number of questions they might ask in the forum, indicating a potential need for guidance on forum etiquette regarding multiple threads.

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[SOLVED] Simple (was thought to be) Gravity Question

Homework Statement


A ball is rolled up a constant slope. After 3.6 s it reaches its maximum displacement of 2.6m, and then begins to roll back down. What was the initial velocity of the ball when it started up the slope?


Homework Equations





The Attempt at a Solution


i know that d=vt of course, but, 2.6=v(3.6), and V would equal 2.6 divided by 3.6 which is .722, and the answer in the key is 1.4m/s?? In all likelihood, its probably something extremely obvious?
 
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You can't use d=vt because there is an acceleration due to gravity
use d=vt-.5at^2
 
you mean d=Vit+.5at^2, emphasis on plus sign?
 
well yes, in this case the acceleration is negative but the formula you gave is the more general form.
 
grantP said:
you mean d=Vit+.5at^2, emphasis on plus sign?

edit: thanks by the way, the question screwed me up w/ the word constant, plus I am taking my very first physics course (phys. 11)----one more question is i might have quite a few questions today...will mods get mad for me making quite a few threads? (maybe 10?)
 
If they're the same types of problems I'd just stick them in the same thread.
 

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