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+(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=(1)/(2)gt^2`

`\Rightarrow g=(2h)/(t^2)\\ \Rightarrow g=(2* 183ft)/((3.5s)^2)=29.87 ft/s^2`

The uncertainty in this value is given by:

`\Delta g=g\sqrt{((\Delta h)/(h))^2+((2\Delta t)/(t))^2}`

Substitute the values:

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