(i) Draw the straight line that best fits the data.

(ii) State why the data do not support the hypothesis.

Another student suggests that the relationship between t and h is of the form
\[t = k\sqrt {\frac{1}{h}} \]
where k is a constant.

To test whether or not the data support this relationship, a graph of t² against \(\frac{1}{h}\) is plotted as shown below.

The best-fit line takes into account the uncertainties for all data points.

The uncertainty in t² for the data point where \(\frac{1}{h} = 10.0{{\rm{m}}^{ - 1}}\) is shown as an error bar on the graph.

(i) State the value of the uncertainty in t² for \(\frac{1}{h} = 10.0{{\rm{m}}^{ - 1}}\).

(ii) Calculate the uncertainty in t² when t = 0.8 ± 0.1s. Give your answer to an appropriate number of significant digits.

(iii) Use the graph to determine the value of k. Do not calculate its uncertainty.

(iv) State the unit of k.

(i) any straight line that goes through all error bars;
(ii) line does not go through origin / (0,0) / zero;

(i) ±0.35s²; (accept answers in range 0.3 to 0.4)

(ii) \(\frac{{\Delta \left( {{t^2}} \right)}}{{{t^2}}} = 2\frac{{\Delta t}}{t}\);
\(\Delta \left( {{t^2}} \right) = {0.8^2} \times 2 \times \frac{{0.1}}{{0.8}}\);
\(\Delta \left( {{t^2}} \right) = 0.16 \approx 0.2{{\rm{s}}^2}\);
answer given to one significant figure;

or

percentage uncertainty in \(t = \frac{{0.1}}{{0.8}} \times 100 = 12.5\% \);
percentage uncertainty in t²=25%;
absolute uncertainty in t=0.25×0.8²=0.16≈0.2s²;
answer given to one significant figure;

(iii)

use of gradient triangle over at least half of line;
value of gradient = 0.30; (accept answers in range 0.28 to 0.32)
=k² to give k=0.55; (accept answers in range 0.53 to 0.57)
or
equation of line is \({t^2} = \frac{{{k^2}}}{h}\);
data values for a point on the line selected;
values substituted into equation to get k=0.55; (accept answers in range 0.53 to 0.57)
Award [2] for answers that use a data point not on the best fit line.

(iv) \({{\rm{m}}^{\frac{1}{2}}}{\rm{s}}\);