Free-Fall Physics Homework: Find Velocity After 2s

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SUMMARY

The discussion centers on calculating the final velocity of a rock dropped from a cliff after 2 seconds, using the equation vf = vi + g∆t. The correct final velocity, accounting for gravity as -9.81 m/s², is -19.6 m/s, which corresponds to option a) -2.0 * 10^1 m/s. The confusion arises from the application of significant figures, where 19.6 m/s can be rounded to 20.0 m/s, but the negative sign indicates the direction of the velocity vector.

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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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