How can I plot more than one value in the same date?

An object of mass m is dropped from a height h above the ground. Plot Y vs X graph.?

• An object of mass m is dropped from a height h above the ground. The speed at which it hits the ground is v. From conservation of energy 1/2mv2 = mgh. You make multiple measurements of h and v. You want to use your h and v values to calculate a value for g. In order to eliminate errors and make outlying data points easily recognizable, you want to obtain g by linearizing your data. That is, you want to make a plot of Y vs. X with Y=gX. g is then given by the slope of a line fitted to your data. Which of the following methods would enable you to obtain g from the slope of the best fit line? Check all that apply. A) A plot of Y vs. X with Y = v2 and X = 2h B) A plot of Y vs. X with Y = v and X = sqrt(2h) C) A plot of Y vs. X with Y = v2/2 and X=h D) A plot of Y vs. X with Y = 2h and X = v2 My answer would be all of the above since they are equivalent. However, I'm a little confused about the last one. Am I allowed to reverse Y and X? Even though the slope should be correct if I were to plot the graph, the value itself wouldn't be correct in the equation above, right? Thanks in advance for any help you can provide!

• Answer:

  Only A) and C) should be used … … ½ m v ² = m g h … --> … ½ v ² = g h = … To be put in the linear form y = g x … A) … y = v ² … x = 2 h … with … y = g x … --> … v ² = g (2 h ) … which reduces to … ½ v ² = g h … same as above … B) … y = v … x = √ [ 2 h ] … with … y = g x … --> … v = g √ [ 2 h ] … which reduces to … v ² = g ² [ 2 h ] ≠ 2 g h … so this should not be used … C) … y = ½ v ² … x = h … with … y = g x … --> … ½ v ² = g h … … same as that from conservation of energy … D) … y = 2 h … x = v ² … with … y = g x … --> … 2h = g v ² … which reduces to … ½ v ² = h / g ≠ g h … so this should not be used … That is, only A) and C) should be used.