Free-Fall Physics Homework: Find Velocity After 2s

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Homework Help Overview

The problem involves calculating the velocity of a rock dropped from a cliff after 2 seconds, specifically within the context of free-fall physics, while ignoring air resistance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of gravitational acceleration and the initial conditions of the problem. There is a focus on the interpretation of the negative sign for gravity and how it affects the final velocity calculation. Questions arise regarding the application of significant figures in the context of the results.

Discussion Status

Some participants have provided insights on the use of negative gravity in the calculations, while others are seeking clarification on how to apply significant figures to the results. Multiple interpretations of the problem are being explored, particularly regarding the final answer and its representation.

Contextual Notes

There is a noted discrepancy between the calculated velocity and the answer provided in the textbook, which prompts further discussion on the assumptions made regarding direction and significant figures.

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Homework Statement



If a boy dropped a rock off a high cliff, how fast would it be falling after 2.0 seconds had elapsed? (ignore air resistance)

a) -2.0 * 10^1 m/s
b) 2.0 * 10^2 m/s
c) 35 m/s
d) -35 m/s

Homework Equations



vf = vi + g∆t

vf = final velocity

vi = initial velocity

g = gravity = -9.81 m/s^2

∆t = change in time

The Attempt at a Solution



since the initial velocity is zero,

the velocity after 2 seconds is

V(t) = 0 + gt,

V(t) = gt,

V(2) = (9.8 m/s2)(2s),

V(2) = 19.6 m/s

The answer in my book says it is a) -2.0 * 10^1 m/s
How do you get that answer?
 
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If you take g negative, V(2) = -gt.
Apply the significant number rule to get the result.
 


rl.bhat said:
If you take g negative, V(2) = -gt.
Apply the significant number rule to get the result.

I'm not too sure on how to apply the signif. number rule; could you please elaborate
 


5minutes said:
I'm not too sure on how to apply the signif. number rule; could you please elaborate

19.6 m/s can be written a 20.0 m/s.
 

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