Chapter 2.2, Problem 21E

To determine

To find: $a,v_0$ and $s_0$ for the given situation.

Answer to Problem 21E

  $\begin{array}{l}a=-32\text{ feet per second square}\\ v_0=138\text{ feet per second}\\ s_0=2\text{ feet}\end{array}$

Explanation of Solution

Given information:

The formula to calculate the distance is as follows:

  $s=\frac{1}{2}a{t}^2+v_{0}t+s_0$

Here,

  $a=$ Acceleration (in feet per Second Square)

  $v_0=$ Velocity (in feet per second)

  $s_0=$ Initial height (in feet)

  $s=$ Height (in feet)

  $t=$ Time (in seconds)

According to the question,

Height $\left(s\right)$	Time $\left(t\right)$
$s_1=124\text{ feet}$	$t_1=1\sec$
$s_2=272\text{ feet}$	$t_2=3\sec$
$s_3=82\text{ feet}$	$t_3=8\sec$

Calculation:

Substitute the given values in the formula and evaluate the system of equations will be as follows:

  $\begin{array}{c}124=\frac{1}{2}a+v_0+s_0\mathrm{...}\text{(1)}\\ 272=\frac{9}{2}a+3v_0+s_0\mathrm{...}\text{(2)}\\ 82=32a+8v_0+s_0\mathrm{...}\text{(3)}\end{array}$

Subtract equation (1) from equation (3) to eliminate $s_0$ variable.

  $27.5a+5v_0=-190\mathrm{...}\left(4\right)$

Now, multiply the above equation by $\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$ to get,

  $38.5a+7v_0=-266\mathrm{...}\left(5\right)$

Subtract equation (2) from equation (3) to eliminate $s_0$ variable.

  $31.5a+7v_0=-42\mathrm{...}\left(6\right)$

Subtract equation (6) from equation (5) to eliminate $v_0$ variable.

  $\begin{array}{c}7a=-224\\ a=\frac{-224}{7}\\ =-32\end{array}$

Put the value of $a$ in equation (6) and evaluate $v_0$ as follows:

  $\begin{array}{c}31.5\left(-32\right)+7v_0=-42\\ -1008+7v_0=-42\\ 7v_0=1008-42\\ 7v_0=966\\ v_0=\frac{966}{7}\\ =138\end{array}$

Put the value of $a$ and $v_0$ in equation (1) and evaluate $s_0$ as follows:

  $\begin{array}{c}124=\frac{1}{2}\left(-32\right)+138+s_0\\ 124=-16+138+s_0\\ s_0=124+16-138\\ =140-132\\ =2\end{array}$

The values of $a,v_0$ and $s_0$ is $-32\text{ feet per second square},138\text{ feet per second}$ and 2 feet respectively.