Question

Suppose you are helping Galileo measure the acceleration due to gravity by dropping a canon ball from the tower of Pisa, You measure the height of the tower to be h =183 ft, and you estimate that the uncertainty is delta h=0.2 ft. You measure the drop time using your pulse and arrive at delta t =3.5 sec with an estimated uncertainty delta t= 0.5 sec. You perform the experiment just one time. If the formula propagated error is

delta g= g (√(delta h/h)^2 +(2* delta t/t)^2)

what would you report as your result (including uncertainty) in ft/sec?

Answer 1

It is given that the height of the tower is

$h=183 ft.$

The uncertainty the measurement of this height is

$\Delta h=0.2 ft$

Drop time is measured as:

$t=3.5s$

The uncertainty in measurement of time is:

$\Delta t=0.5 s$

Using the equation of motion: $h=ut+\frac{1}{2} at^2$ where, $h$ is the distance covered, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time.

$u=0$ (because canon ball is in free fall). we need to calculate the value of a=g.

$\Rightarrow h=\frac{1}{2}gt^2$

$\Rightarrow g=\frac{2h}{t^2}\\ \Rightarrow g=\frac{2\times 183ft}{(3.5s)^2}=29.87 ft/s^2$

The uncertainty in this value is given by:

$\Delta g=g\sqrt{(\frac{\Delta h}{h})^2+(\frac{2\Delta t}{t})^2}$

Substitute the values:

$\Delta g=29.87\sqrt{(\frac{0.2 }{183})^2+(\frac{2\times 0.5}{3.5})^2}=29.87\sqrt{1.19\times 10^{-6}+0.08}=29.87\times \sqrt{0.08}=29.87\times 0.28=8.44 ft/s^2$