Chapter 4.8, Problem 46E

To determine

whether the balls ever pass each other.

Answer to Problem 46E

The balls pass each other at the time of $\underset{\_}{t=5\text{seconds}}$.

Explanation of Solution

Given data:

Consider the edge of cliff is at 432 ft above the ground.

Speed of first ball is $48\frac{\text{ft}}{\text{s}}$ and speed of second ball is $24\frac{\text{ft}}{\text{s}}$.

Formula Used:

Write the expression for acceleration function $a\left(t\right)$.

$a\left(t\right)=v^{\prime }\left(t\right)$ (1)

Here,

$v^{\prime }\left(t\right)$ is first derivative of velocity function $v\left(t\right)$.

Write the expression for velocity function $v\left(t\right)$.

$v\left(t\right)=s^{\prime }\left(t\right)$ (2)

Here,

$s^{\prime }\left(t\right)$ is first derivative of displacement function $s\left(t\right)$.

Antiderivative of $t$ is $\frac{1}{2}{t}^2$ and 1 is $t$.

Calculation:

Initial velocity of first ball is $48\frac{\text{ft}}{\text{s}}$, that is $v_1\left(0\right)=48\frac{\text{ft}}{\text{s}}$ and second ball is thrown after a second with velocity of $24\frac{\text{ft}}{\text{s}}$ , that is at $v_2\left(1\right)=24\frac{\text{ft}}{\text{s}}$.

The two balls are thrown from same edge and edge of cliff is at 432 ft. Hence,

$\begin{array}{c}{s}_1\left(0\right)=s_2\left(0\right)\\ =432\end{array}$

The motion of stone is close to ground, so the motion is considered as gravitational constant $\left(g\right)$, which is $32\frac{\text{ft}}{{\text{s}}^{\text{2}}}$.

Write the expression for acceleration function of first ball $\left(a\left(t\right)\right)$.

$a_1\left(t\right)=-32$

Substitute ${v^{\prime }}_1\left(t\right)$ for $v\left(t\right)$ and $-32$ for $a\left(t\right)$ in equation (1),

${v^{\prime }}_1\left(t\right)=-32$

Antiderivate the expression with respect to t,

$v_1\left(t\right)=-32t+C$ (3)

Here,

C is arbitrary constant.

Substitute 0 for t in equations (3),

$\begin{array}{c}{v}_1\left(0\right)=-32\left(0\right)+C\\ =C\end{array}$

Substitute 48 for $v_1\left(0\right)$,

$C=48$

Substitute 48 for C in equation (3),

$v_1\left(t\right)=-32t+48$ (4)

Substitute $-32t+48$ for $v\left(t\right)$ in equation (2),

${s^{\prime }}_1\left(t\right)=-32t+48$

Antiderivate the expression with respect to t,

$s_1\left(t\right)=-32\left(\frac{1}{2}{t}^2\right)+48t+D$

$s_1\left(t\right)=-16t^2+48t+D$ (5)

Here,

D is arbitrary constant.

Substitute 0 for t in equation (5),

$\begin{array}{c}{s}_1\left(0\right)=-16{\left(0\right)}^2+48\left(0\right)+D\\ =0+D\\ =D\end{array}$

Substitute 432 for $s_1\left(0\right)$,

$D=432$

Substitute 432 for D in equation (5),

$s_1\left(t\right)=-16t^2+48t+432$

Write the expression for acceleration function of second ball $\left(a_2\left(t\right)\right)$.

$a_2\left(t\right)=-32$

Substitute ${v^{\prime }}_2\left(t\right)$ for $v\left(t\right)$ and $-32$ for $a\left(t\right)$ in equation (1),

${v^{\prime }}_2\left(t\right)=-32$

Antiderivate the expression with respect to t,

$v_2\left(t\right)=-32t+C$ (6)

Substitute 1 for t in equations (6),

$\begin{array}{c}{v}_2\left(1\right)=-32\left(1\right)+C\\ =-32+C\end{array}$

Substitute 24 for $v_2\left(1\right)$,

$\begin{array}{l}24=-32+C\\ C=24+32\\ C=56\end{array}$

Substitute 56 for C in equation (6),

$v_1\left(t\right)=-32t+48$ (7)

Substitute $-32t+48$ for $v\left(t\right)$ in equation (2),

${s^{\prime }}_2\left(t\right)=-32t+56$

Antiderivate the expression with respect to t,

$s_2\left(t\right)=-32\left(\frac{1}{2}{t}^2\right)+56t+D$

$s_2\left(t\right)=-16t^2+56t+D$ (8)

Substitute 1 for t in equation (8),

$\begin{array}{c}{s}_2\left(1\right)=-16{\left(1\right)}^2+56\left(1\right)+D\\ =-16+56+D\\ =40+D\end{array}$

Substitute 432 for $s_2\left(1\right)$,

$\begin{array}{l}432=40+D\\ D=432-40\\ D=392\end{array}$

Substitute 392 for D in equation (8),

$s_2\left(t\right)=-16t^2+56t+392$

The balls pass each other when their positions are equal.

$s_1\left(t\right)=s_2\left(t\right)$

Substitute $-16t^2+48t+432$ for $s_1\left(t\right)$ and $-16t^2+56t+392$ for $s_2\left(t\right)$,

$\begin{array}{l}-16t^2+48t+432=-16t^2+56t+392\\ -16t^2+48t+432+16t^2-56t-392=0\\ -8t+40=0\\ 8t=40\end{array}$

$\begin{array}{l}t=\frac{40}{8}\text{seconds}\\ t=5\text{seconds}\end{array}$

Thus, the balls pass each other at the time of $\underset{\_}{t=5\text{seconds}}$.