Chapter 1.2, Problem 64STP

To determine

To find: the distance in feet that an object falls in time seconds using the given expression.

Answer to Problem 64STP

Option $D$ is correct.

Explanation of Solution

Given:

Distance of an object that falls in $t$ seconds is given by an expression:

  $16t^2$ .

Concept used:

Distance or range formula:

  $s=ut+\frac{1}{2}a{t}^2$ .

If something is drop from certain height then the initial velocity is $0$ .

  $s=\frac{1}{2}g{t}^2$ .

here $g$ is an acceleration due to Gravity.

By putting the value of time the equation can be solved.

Calculation:

Distance of an object that falls in $t$ seconds is given by an expression:

  $16t^2$ .

Where distance is calculated in terms of feet.

Expression for distance, if the ball is dropped from a building in $3s$ is given by:

  $16t^2$ where $t=3$ :

Put $t=3$ , in given expression, it becomes:

  $\begin{array}{l}\Rightarrow 16\times 3^2.\\ \Rightarrow 16\times 9.\\ \Rightarrow 144.\end{array}$

So, the ball is dropped from a tall building in $3s$ , the distance will be $144\text{ft}$ .

Hence, option $D$ is correct.