Why doesn't my result yield 9.8 m/s^2?

  • Thread starter Henrybar
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  • #1
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A ball falling from rest is located 45meters below its starting point 3.0s later. Assuming its acceleration is uniform, what is it's value? (air resistance is negligible)

Relevant Equations - Equations for uniformly accelerated motion
Δd=(v1 x Δt)+(a x Δt^2)/2

Attempt To solve:
v1= 0
Δd=45m
Δt=3.0s
a=?

a=(2Δd/Δt^2)-(2v1Δt/Δt^2)
a= ((2 x 45)/9)-((0 x 3)/9)
a=10m/s^2

If a ball is falling from rest(0 initial velocity) shouldn't it's acceleration be 9.8 seconds since gravity is constant? Is my equation wrong? Is my idea of time-acceleration misplaced?
 

Answers and Replies

  • #2
gneill
Mentor
20,925
2,867
The given values result in the acceleration that you calculated. Your calculation is okay.

In introductory classes g is often approximated to be 10 m/s2 because it simplifies numerical calculations while still getting across the concepts being taught.
 
  • #3
AlephZero
Science Advisor
Homework Helper
7,002
293
The question doesn't say the ball was near the surface of the earth. Don't make unjustified assumptions! :smile:
 

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