Evaluating minimum and maximum values with calculations

[crememars]

Homework Statement

Determine the maximum and minimum values for the average speed of an object falling from a height of 2 meters in 0.63 seconds.

Uncertainty of time = +/- 0.05 s
Uncertainty of height = +/- 0.02 m

Relevant Equations

v = ∆y/∆t

Hi! I know this may seem like a silly question but I really just want to make sure I understand this correctly. I've already calculated the minimum and maximum values for time and height:

t min = 0.58 s
t max = 0.68 s
y min = 1.98 m
y max = 2.02 m

To calculate the minimum average speed, would I use the minimum values of time and height ? (and vice versa for max speed)
I thought that would be the process, but the minimum values give a bigger speed, and the max values give me a smaller speed:

v measured = 2.00/0.63 = 3.17 m/s
v min = 1.98/0.58 = 3.41 m/s
v max = 2.02/0.68 = 2.97 m/s

This is really confusing..

 

[erobz]

If you want $\bar{v}_{min}$ you have to choose the uncertainties such that the quantity: $$ \bar{v} = \frac{ h \pm u_h}{t \pm u_t}$$ is as small as it can be.

How would you choose the signs to do that?

 

[hmmm27]

v = ∆y/∆t .

 

[crememars]

If you want $\bar{v}_{min}$ you have to choose the uncertainties such that the quantity: $$ \bar{v} = \frac{ h \pm u_h}{t \pm u_t}$$ is as small as it can be.

How would you choose the signs to do that?

addition gives the smallest value.. so I'd essentially be using the maximum values to get a minimum value for speed? that's very weird haha

 

[erobz]

addition gives the smallest value.. so I'd essentially be using the maximum values to get a minimum value for speed? that's very weird haha

You have you think about the relative sizes of the numerator and denominator. Think about what happens in the division. Addition is not the answer, it depends on which part (numerator, denominator) you are talking about?

 

[crememars]

You have you think about the relative sizes of the numerator and denominator.

Wait, would it be correct to divide the minimum distance by the maximum time then? Ohh it makes a bit more sense now, I see how that would produce a minimum speed