Chapter 2, Problem 84GP

To determine

The difference in the estimates of building height in the given time.

Answer to Problem 84GP

6.3 m

Explanation of Solution

Given:

Initial velocity of the barometer as it is dropped $=v_o=0\frac{\text{m}}{\text{s}}$

Time of travel noted by one student for barometer $=t_1=2\sec$

Time of travel noted by other student for barometer $=t_2=2.3\sec$

Height of the building as noted by one student $=h_1$

Height of the building as noted by other student $=h_2$

Vertical displacement of barometer as noted by one student $=\Delta y_1=h_1$

Vertical displacement of barometer as noted by other student $=\Delta y_2=h_2$

Acceleration experienced by the barometer during free fall $=a=g=9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Difference in the heights of building noted by two students $=\Delta h$

Formula Used:

Displacement of an object in constant accelerated motion for a given time interval is given as

  $\Delta y=v_{o}t+(0.5)a{t}^2$

Calculation:

Consider the motion of the barometer as observed by one student.

Displacement of the barometer is given as

  $\Delta y_1=v_o{t}_1+(0.5)a{t}_1{}^2$

Substituting the values

  $\begin{array}{l}{h}_1=(0)(2)+(0.5)(9.8){(}^2\\ h_1=19.6\text{m}\end{array}$

Consider the motion of the barometer as observed by other student.

Displacement of the barometer is given as

  $\Delta y_2=v_o{t}_2+(0.5)a{t}_2{}^2$

Substituting the values

  $\begin{array}{l}{h}_2=(0)(2.3)+(0.5)(9.8){(}^{2.3}\\ h_2=25.92\text{m}\end{array}$

Difference in the heights of building is given as

  $\begin{array}{l}\Delta h=h_2-h_1\\ \Delta h=25.92-19.6\\ \Delta h=6.3\text{m}\end{array}$

Conclusion:

Therefore, the difference in the estimates of height is $6.3\text{m}$