Chapter 4, Problem 13A

How far will a freely falling object fall from rest in five seconds? Six seconds?

To determine

The distance which a freely falling object travels from a position of rest for a given time duration.

Answer to Problem 13A

The distance travelled by the object in 5 s is $122.5\text{}\text{m}$.

The distance travelled by the object in 6 s is $176.4\text{}\text{m}$.

Explanation of Solution

Given:

A freely falling object initially at rest travels for time instants 5 s and 6 s

Formula used:

The equation of motion which determines the distance travelled or displacement of body with an initial velocity and moving with a particular value of acceleration for a given time is,

  $s=ut+\frac{1}{2}a{t}^2$

In this case, a freely falling object is considered. This implies the acceleration acting on it is only acceleration due to gravity, g. The force due to air resistance is also neglected. Thus, the above equation can be rewritten as,

  $s=ut+\frac{1}{2}g{t}^2$ ……. (1)

The value of acceleration due to gravity on earth is $g=9.8{\text{ m/s}}^2$

Calculation:

Since the object starts from rest, its initial velocity is $u=0$ . Now, the distance travelled by the object in 5 seconds can be calculated by substituting it along with the acceleration due to gravity in (1)

  $\begin{array}{c}s=\frac{1}{2}\times 9.8\times {\left(5\right)}^2\\ =122.5\text{ m}\end{array}$

Considering the next case when time of travel is given as t =6 s, the distance travelled by this time is,

  $\begin{array}{c}s=\frac{1}{2}\times 9.8\times {\left(6\right)}^2\\ =176.4\text{ m}\end{array}$

Conclusion:

The displacement of the object from its initial position or the distance travelled in 5 seconds is 122.5 m and in 6 seconds is 176.4 m.