Chapter 4.8, Problem 43E

(a)

To determine

To find: The distance of stone above the ground level.

Answer to Problem 43E

The distance of stone above ground level at time t is $\underset{\_}{450-4.9t^2}$.

Explanation of Solution

Given data:

The distance between upper observation deck and ground is 450 m.

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 position function $s\left(t\right)$.

Antiderivative of $t^3$ is $\frac{1}{4}{t}^4$, $t^2$ is $\frac{1}{3}{t}^3$, $t$ is $\frac{1}{2}{t}^2$ and 1 is $t$.

Calculation:

At first observation, velocity is zero and position is at 450 m at $t=0$, as the stone is dropped 450 m height to ground level. Hence,

$\begin{array}{l}v\left(0\right)=0\\ s\left(0\right)=450\end{array}$

The motion of stone is close to ground, so the motion is considered as gravitational constant $\left(g\right)$, which is $9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}$.

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

$a\left(t\right)=-9.8$

Substitute $-9.8$ for $a\left(t\right)$ in equation (1),

$v^{\prime }\left(t\right)=-9.8$

Antiderivate the expression with respect to t,

$v\left(t\right)=-9.8t+C$ (3)

Here,

C is arbitrary constant.

Substitute 0 for t in equations (3),

$v\left(0\right)=-9.8\left(0\right)+C$

$v\left(0\right)=C$ (4)

Substitute 0 for $v\left(0\right)$ in equation (4),

$C=0$

Substitute 0 for C in equation (3),

$v\left(t\right)=-9.8t+0$

$v\left(t\right)=-9.8t$ (5)

Substitute $-9.8t$ for $v\left(t\right)$ in equation (2),

$s^{\prime }\left(t\right)=-9.8t+C$

Antiderivate the expression with respect to t.

$s\left(t\right)=-9.8\left(\frac{1}{2}{t}^2\right)+D$

$s\left(t\right)=-4.9t^2+D$ (6)

Here,

D is arbitrary constant.

Substitute 0 for t in equation (6),

$\begin{array}{c}s\left(0\right)=-4.9\left(0^2\right)+D\\ =0+D\\ =D\end{array}$

Substitute 450 for $s\left(0\right)$,

$D=450$

Substitute 450 for D in equation (6),

$\begin{array}{c}s\left(t\right)=-4.9t^2+450\\ =450-4.9t^2\end{array}$

Thus, the distance of stone above ground level at time t is $\underset{\_}{450-4.9t^2}$.

(b)

To determine

To find: The time that stone takes to reach ground.

Answer to Problem 43E

The time that stone takes to reach ground is 9.58 seconds.

Explanation of Solution

Given data:

The distance between upper observation deck and ground is 450 m.

Calculation:

When the stone reaches the ground, the positional function reaches zero. Hence,

$s\left(t\right)=0$ (7)

Substitute $450-4.9t^2$ for $s\left(t\right)$ in equation (7),

$\begin{array}{l}450-4.9t^2=0\\ 4.9t^2=450\\ t^2=\frac{450}{4.9}\\ t^2=91.84\end{array}$

Take square root on both sides.

$\begin{array}{l}t=\sqrt{91.84}\text{seconds}\\ t\simeq 9.58\text{seconds}\end{array}$

Thus, the time that stone takes to reach ground is 9.58 seconds.

(c)

To determine

To find: The velocity that stone strikes the ground.

Answer to Problem 43E

The velocity that stone strikes the ground is $\underset{\_}{-93.9\frac{\text{m}}{\text{s}}}$.

Explanation of Solution

Given data:

The distance between upper observation deck and ground is 450 m.

Calculation:

Substitute 9.58 seconds for t in equation (5),

$\begin{array}{c}v\left(t\right)=-\left(9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\left(9.58\text{seconds}\right)\\ =-93.88\frac{\text{m}}{\text{s}}\\ \simeq -93.9\frac{\text{m}}{\text{s}}\end{array}$

Thus, the velocity that stone strikes the ground is $\underset{\_}{-93.9\frac{\text{m}}{\text{s}}}$.

(d)

To determine

To find: The time that stone takes to reach ground.

Answer to Problem 43E

The time that stone takes to reach ground, when it is thrown downward with speed of $5\frac{\text{m}}{\text{s}}$ is 9.09 seconds.

Explanation of Solution

Given data:

The speed is $5\frac{\text{m}}{\text{s}}$.

Formula used:

Write the expression to find the roots of $a{t}^2+bt+c=0$.

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ (8)

Here,

a, b, and c are constants.

Calculation:

The stone is thrown downwards with a speed of $5\frac{\text{m}}{\text{s}}$. Hence $v\left(0\right)=-5$.

Substitute –5 for $v\left(0\right)$ in equation (4),

$C=-5$

Substitute –5 for C in equation (3),

$\begin{array}{c}v\left(t\right)=-9.8t-5\\ =-\left(9.8t+5\right)\end{array}$

Substitute $-\left(9.8t+5\right)$ for $v\left(t\right)$ in equation (2),

$s^{\prime }\left(t\right)=-\left(9.8t+5\right)+C$

Antiderivate the expression with respect to t,

$s\left(t\right)=-9.8\left(\frac{1}{2}{t}^2\right)-5t+D$

$s\left(t\right)=-4.9t^2-5t+D$ (9)

Here,

D is arbitrary constant.

Substitute 0 for t in equation (9),

$\begin{array}{c}s\left(0\right)=-4.9\left(0^2\right)-5\left(0\right)+D\\ =0+D\\ =D\end{array}$

Substitute 450 for $s\left(0\right)$,

$D=450$

Substitute 450 for D in equation (9),

$\begin{array}{c}s\left(t\right)=-4.9t^2-5t+450\\ =450-4.9t^2-5t\end{array}$

Substitute $450-4.9t^2-5t$ for $s\left(t\right)$ in equation (7),

$\begin{array}{l}450-4.9t^2-5t=0\\ 4.9t^2+5t-450=0\end{array}$

Compare the expressions $a{x}^2+bx+c=0$ and $4.9t^2+5t-450=0$.

$\begin{array}{l}a=4.9\\ b=5\\ c=-450\end{array}$

Substitute 4.9 for a, 5 for b, and –450 for c in equation (8),

$\begin{array}{c}t=\frac{-5\pm \sqrt{5^2-4\left(4.9\right)\left(-450\right)}}{2\left(4.9\right)}\\ =\frac{-5\pm \sqrt{25+8820}}{9.8}\\ =\frac{-5\pm \sqrt{8845}}{9.8}\\ =\frac{-5\pm 94.05}{9.8}\end{array}$

Hence, the two possible values of t are,

$\begin{array}{c}t=\frac{-5+94.05}{9.8}\\ =\frac{89.05}{9.8}\\ =9.086\\ \simeq 9.09\text{seconds}\end{array}$

$\begin{array}{c}t=\frac{-5-94.05}{9.8}\\ =\frac{-99.05}{9.8}\\ =-10.01\text{seconds}\end{array}$

The time cannot be a negative, so the value of t is 9.09 seconds.

Thus, the time that stone takes to reach ground, when it is thrown downward with speed of $5\frac{\text{m}}{\text{s}}$ is 9.09 seconds.