ay=-9.8m/s^2 v1y=0m/s ∆dy=-91.2cm (-0.912m)±1.5mm (0.0015m) ∆t=?
∆dy= v1y∆t + 1/2ay∆t^2
-0.912m±0.0015m=0∆t + 1/2(-9.8m/s2)∆t^2
-0.912m±0.0015m=-4.9m/s^2∆t^2
0.1861224499s2±0.16%=∆t2
√0.186122449s2±0.16%=√∆t^2
0.43141911s±0.4%=∆t
0.43s±0s=∆t
is this correct?

Question

ay=-9.8m/s^2        v1y=0m/s        ∆dy=-91.2cm (-0.912m)±1.5mm (0.0015m)         ∆t=?
∆dy= v1y∆t + 1/2ay∆t^2
-0.912m±0.0015m=0∆t + 1/2(-9.8m/s2)∆t^2
-0.912m±0.0015m=-4.9m/s^2∆t^2
0.1861224499s2±0.16%=∆t2
√0.186122449s2±0.16%=√∆t^2
0.43141911s±0.4%=∆t
0.43s±0s=∆t
is this correct?

anonymous · Answer

let me see
V=da/dy=-9.8 m/ss*t =vly=0
y(t)=dV/dt=-1/2*9.8m/ss t^2=-91.2cm
then t^2=-91.2cm*2/-9.8m=>0.186122
t=0.431419 
then check -.5*9.8*t^2=0.912 m works

anonymous · Answer

okay thanks! but are my uncertainty calculations correct though? Im not sure if I'm suppose to square root the 0.16% too

anonymous · Answer

Those can be tricky, you were probably given an equation to use for them that is less complex then what you would use in research.

anonymous · Answer

http://en.wikipedia.org/wiki/Propagation_of_uncertainty link to the equation for error prorogation.. is this what you are working with?

anonymous · Answer

Like i know when you add or subtract data you would add absolout uncertanties, and when you multiply or divide data you would add the relative percentage uncertainties, so I'm so quite sure what you do when you need to square root an uncertainty

anonymous · Answer

no I'm not working with error prorogation

anonymous · Answer

it not normally that easy, if that is what your teacher wants, they that is easy. else is is probably something like new error=1/2(olderror/number)

anonymous · Answer

If you have not done anything like that then you are probably doing something different. These equations come form standard deviation.

anonymous · Answer

Most of the time a computer will do them for you.

anonymous · Answer

okay thanks anyways