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

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?

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