Chapter 3, Problem 81A

To determine

To Compare: The displacements of two objects during the same time interval.

Answer to Problem 81A

  $s_1=s_2=240.1\text{ m}$

Explanation of Solution

Given:

Time interval, $t=7.0\text{ s}$

Acceleration of first object due to gravity, $g=-9.8{\text{ m/s}}^2$

Acceleration of second object due to gravity, $g=9.8{\text{ m/s}}^2$

Formula used:

First equation of kinematics:

  $v=u+at$

Here, v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.

Calculation:

Calculation for first object:

At maximum height final velocity of first object would be zero.

Let initial velocity of first object be $u$

From first equation of kinematics:

  $\begin{array}{c}v=u+at\\ 0=u-9.8\times 7.0\\ u=68.6\text{ m/s}\end{array}$

Displacement of first object can be calculated from third equation of kinematics:

  $\begin{array}{c}{v}^2=u^2-2g{s}_1\\ \therefore \\ s_1=\frac{v^2-u^2}{-2g}\\ =\frac{0^2-{\left(68.6\right)}^2}{-2\times 9.8}\\ =240.1\text{ m}\end{array}$

Calculation for second object:

Initial velocity of second object would be zero.

Final velocity of second object can be calculated from third equation of kinematics:

  $\begin{array}{c}v=u+gt\\ =0+9.8\times 7.0\\ =68.6\text{ m/s}\end{array}$

Displacement of second object:

  $\begin{array}{c}{s}_2=\frac{v^2}{2g}\\ =\frac{{\left(68.6\right)}^2}{2\times 9.8}\\ =240.1\text{ m}\end{array}$

Conclusion:

Displacement of both objects would be same in same interval of time.