Question

Two students are asked to find the height of a particular building using a barometer. Instead of using the barometer as an altitude measuring device, they take it to the roof of the building and drop it off, timing its fall. One student reports a fall time of 2.0 s, and the other, 2.3 s. What % difference does the 0.3 s make for the estimates of the building’s height?

Short answer

The obtained percentage difference in the heights of the building is \(27.7\% \).

Step-by-step solution

Kinematics equation

In solving problems related to percentage, first, calculate the difference in the heights of the building reported by the two students and then divide it with the average height of the building.

Given data:

The fall time reported by one student is\({t_1} = 2\;{\rm{s}}\).

The fall time reported by the other student is\({t_2} = 2.3\;{\rm{s}}\).

The relation used to find the height of the building is given by:

\(h = u{t_1} + \frac{1}{2}g{t_1}^2\)

Here, u is the initial speed whose value is zero as measured by the barometer, and g is the gravitational acceleration.

On plugging the values in the above relation, you get:

\(\begin{aligned}h &= \left( 0 \right){t_1} + \frac{1}{2}\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right){\left( {2\;{\rm{s}}} \right)^2}\\h &= 19.62\;{\rm{m}}\end{aligned}\)

Calculation of the height of the building

The relation used to find the height of the building as reported by the other student is given by:

\(h' = u{t_2} + \frac{1}{2}g{t_2}^2\)

On plugging the values in the above relation, you get:

\(\begin{aligned}h' &= \left( 0 \right){t_2} + \frac{1}{2}\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right){\left( {2.3\;{\rm{s}}} \right)^2}\\h' &= 25.94\;{\rm{m}}\end{aligned}\)

Calculation of percentage difference in building’s height

The relation used to find the average height of the building is given by:

\(\begin{aligned}H &= \frac{{h + h'}}{2}\\H &= \left( {\frac{{19.62\;{\rm{m}} + 25.94\;{\rm{m}}}}{2}} \right)\\H &= 22.78\;{\rm{m}}\end{aligned}\)

To determine the percentage, you can use the formula

\(\% P = \frac{{h' - h}}{H} \times 100\% \)

On plugging the values in the above relation, you get:

\(\begin{aligned}\% P &= \frac{{25.94\;{\rm{m}} - 19.62\;{\rm{m}}}}{{22.78\;{\rm{m}}}} \times 100\% \\\% P &= 27.7\% \end{aligned}\)

Thus, the percentage difference is \(27.7\% \).