Chapter 4, Problem 78PQ

(a)

To determine

Find the horizontal displacement of the dart as a function of initial speed.

Answer to Problem 78PQ

The horizontal displacement of the dart as a function of initial speed is $x\left(v_0\right)=0.0508v_0^2+0.743v_0\sqrt{0.00466v_0^2+0.214}$.

Explanation of Solution

Write the kinematic equation.

    $y_f=y_i+v_{y0}t+\frac{1}{2}a{t}^2$                                                             (I)

Here, $y_f$ is the y component of final position, $y_i$ is the y component of initial position, $v_{yo}$ is the y component of initial speed, $a$ is the acceleration and $t$ is the time.

    $x_f=x_i+v_{x0}t+\frac{1}{2}{a}_x{t}^2$                                                             (II)

Here, $x_f$ is the x component of final position, $x_i$ is the x component of initial position, $v_{xo}$ is the x component of initial speed, $a_x$ is the x component of acceleration and $t$ is the time.

Conclusion:

Substitute $\text{1}\text{.05 m}$ for $y_i$, $v_0\sin 42.0°$ for $v_{y0}$ and $\text{4}{\text{.91 m/s}}^2$ for $a$ in equation I.

    $y_f=\text{1}\text{.05 m}+\left(v_0\sin 42.0°\right)t+\frac{1}{2}\left(\text{4}{\text{.91 m/s}}^2\right)t^2$

The above equation is a quadratic and the solution for time will be,

    $t=\frac{0.670v_0\pm \sqrt{{\left(0.670v_0\right)}^2+4\left(4.91\right)\left(1.05\right)}}{9.80}$

Since time should be position take the positive root then the value of time is,

    $t=0.0683v_0+\sqrt{0.00466v_0^2+0.214}$

Substitute $0$ for $x_i$, $v_0\cos 42.0°$ for $v_{x0}$, $0.0683v_0+\sqrt{0.00466v_0^2+0.214}$ for $t$ and $0$ for $a_x$ in equation II.

    $\begin{array}{c}{x}_f=0+\left(v_0\cos 42.0°\right)\left(0.0683v_0+\sqrt{0.00466v_0^2+0.214}\right)+\frac{1}{2}\left(0\right)t^2\\ x\left(v_0\right)=0.0508v_0^2+0.743v_0\sqrt{0.00466v_0^2+0.214}\end{array}$

Therefore, the horizontal displacement of the dart as a function of initial speed is $x\left(v_0\right)=0.0508v_0^2+0.743v_0\sqrt{0.00466v_0^2+0.214}$.

(b)

To determine

Find the horizontal displacement with initial velocity of $0.300\text{ m/s}$.

Answer to Problem 78PQ

The horizontal displacement with initial velocity of $0.300\text{ m/s}$ is $\text{0}\text{.108 m}$.

Explanation of Solution

Use the horizontal displacement equation derived in previous part.

    $x\left(v_0\right)=0.0508v_0^2+0.743v_0\sqrt{0.00466v_0^2+0.214}$                                                    (III)

Conclusion:

Substitute $0.300\text{ m/s}$ for $v_0$ in equation III.

    $\begin{array}{c}x\left(v_0=0.300\right)=0.0508{\left(0.300\right)}^2+0.743\left(0.300\right)\sqrt{0.00466{\left(0.300\right)}^2+0.214}\\ =0.108\text{ m}\end{array}$

Therefore, the horizontal displacement with initial velocity of $0.300\text{ m/s}$ is $\text{0}\text{.108 m}$.

(c)

To determine

Find the horizontal displacement with initial velocity of $30.0\text{ m/s}$.

Answer to Problem 78PQ

The horizontal displacement with initial velocity of $30.0\text{ m/s}$ is $\text{92}\text{.4 m}$.

Explanation of Solution

Use the horizontal displacement equation derived in previous part.

Conclusion:

Substitute $30.0\text{ m/s}$ for $v_0$ in equation III.

    $\begin{array}{c}x\left(v_0=30.0\right)=0.0508{\left(30.0\right)}^2+0.743\left(30.0\right)\sqrt{0.00466{\left(30.0\right)}^2+0.214}\\ =92.4\text{ m}\end{array}$

Therefore, the horizontal displacement with initial velocity of $30.0\text{ m/s}$ is $\text{92}\text{.4 m}$.

(d)

To determine

Drive the simplified expression for horizontal displacement for small initial velocity.

Answer to Problem 78PQ

The simplified expression for horizontal displacement for small initial velocity is $0.344v_0$.

Explanation of Solution

When the initial velocity is small then the square of the initial velocity is negligible then equation III becomes.

    $\begin{array}{c}x\left(v_0\right)\cong 0+0.743v_0\sqrt{0+0.214}\\ =0.344v_0\end{array}$

Therefore, the simplified expression for horizontal displacement for small initial velocity is $0.344v_0$.

(e)

To determine

Drive the expression for horizontal displacement for high initial velocity.

Answer to Problem 78PQ

The expression for horizontal displacement for high initial velocity is $0.102v_0^2$.

Explanation of Solution

When the initial velocity is high then the $v_0$ value is negligible comparing to $v_0^2$ then equation III becomes.

    $\begin{array}{c}x\left(v_0\right)\cong 0.0508v_0^2+0.743v_0\sqrt{0.00466v_0^2+0}\\ =0.102v_0^2\end{array}$

Therefore, the expression for horizontal displacement for high initial velocity is $0.102v_0^2$.