Chapter 4, Problem 43AP

A spring cannon is located at the edge of a table that is 1.20 m above the floor. A steel ball is launched from the cannon with speed v_i at 35.0° above the horizontal. (a) Find the horizontal position of the ball as a function of v_i at the instant it lands on the floor. We write this function as x(v_i). Evaluate x for (b) v_i = 0.100 m/s and for (c) v_i = 100 m/s. (d) Assume v_i is close to but not equal to zero. Show that one term in the answer to part (a) dominates so that the function x(v_i) reduces to a simpler form. (c) If v_i is very large, what is the approximate form of x(v)? (f) Describe the overall shape of the graph of the function x(v_i).

To determine

The ball’s horizontal position as a function of $v_{\text{i}}$.

Answer to Problem 43AP

The horizontal position of the ball as a function of $v_{\text{i}}$ is $v_{\text{i}}{\left(0.002299v_{\text{i}}{}^2+0.1643\right)}^{\frac{1}{2}}+0.04794v_{\text{i}}{}^2$.

Explanation of Solution

The location of the spring cannon is $1.20\text{m}$ above the floor and the launch angle of the ball with speed $v_{\text{i}}$ is $35.0°$ above the horizontal.

Write the formula to calculate the vertical distance covered by the ball

    $y=y_{\text{o}}+v_{\text{y}}t-\frac{1}{2}g{t}^2$                                                      (I)

Here, $y$ is the vertical distance covered by the ball, $v_{\text{y}}$ is the component of the velocity in y direction, $t$ is the time interval, $g$ is the acceleration due to gravity and $y_{\text{o}}$ is the initial vertical distance above the floor.

Write the formula to vertical component of the velocity

    $v_{\text{y}}=v_{\text{i}}\sin \theta$

Here, $v_{\text{i}}$ is the velocity of the ball during projectile motion.

Substitute $v_{\text{i}}\sin \theta$ for $v_{\text{i}}$ in equation (I).

    $y=y_{\text{o}}+v_{\text{i}}\sin \theta \times t-\frac{1}{2}g{t}^2$

Substitute $1.2\text{m}$ for $y_{\text{o}}$, $0$ for $y$, $35.0°$ for $\theta$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ to find $t$.

    $\begin{array}{c}0=1.2\text{m}+v_{\text{i}}\sin 35.0°\times t-\frac{1}{2}9.81\text{m}/{\text{s}}^{\text{2}}\times t^2\\ 0=9.81\text{m}/{\text{s}}^{\text{2}}\times t^2-2v_{\text{i}}\sin 35.0°\times t-1.2\text{m}\end{array}$                                              (II)

Solve the equation (II).

    $t=0.05846v_{\text{i}}+{\left(3.4184\times {10}^{-3}{v}_{\text{i}}{}^2+0.2446\right)}^{\frac{1}{2}}$

Write the formula to calculate the horizontal distance covered by the ball

    $x=v_{\text{x}}t$                                                                                            (III)

Here, $v_{\text{x}}$ is the component of velocity in x direction and $t$ is the time interval.

Write the expression for the horizontal component of the velocity

    $v_{\text{x}}=v_{\text{i}}\cos \theta$

Substitute $v_{\text{i}}\cos \theta$ for $v_{\text{x}}$ and $0.05846v_{\text{i}}+{\left(3.4184\times {10}^{-3}{v}_{\text{i}}{}^2+0.2446\right)}^{\frac{1}{2}}$ for $t$ in equation (III).

    $x=v_{\text{i}}\cos \theta \times \left(0.05846v_{\text{i}}+{\left(3.4184\times {10}^{-3}{v}_{\text{i}}{}^2+0.2446\right)}^{\frac{1}{2}}\right)$

Conclusion:

Substitute $35.0°$ for $\theta$ in above expression.

    $\begin{array}{c}x=v_{\text{i}}\cos 35.0°\times \left(0.05846v_{\text{i}}+{\left(3.4184\times {10}^{-3}{v}_{\text{i}}{}^2+0.2446\right)}^{\frac{1}{2}}\right)\\ =v_{\text{i}}{\left(0.002293v_{\text{i}}{}^2+0.1641\right)}^{\frac{1}{2}}+0.04788v_{\text{i}}{}^2\end{array}$                    (IV)

Therefore, the horizontal position of the ball as a function of $v_1$ is $v_{\text{i}}\left(0.002293v_{\text{i}}{}^2+0.164\right)1^{\frac{1}{2}}+0.04788v_{\text{i}}{}^2$.

To determine

The horizontal position of the ball with $v_{\text{i}}=0.100\text{m}/\text{s}$.

Answer to Problem 43AP

The horizontal position the ball with $v_{\text{i}}=0.100\text{m}/\text{s}$ is $0.0410\text{m}$.

Explanation of Solution

From equation (IV),

    $x=v_{\text{i}}{\left(0.002293v_{\text{i}}{}^2+0.1641\right)}^{\frac{1}{2}}+0.04788v_{\text{i}}{}^2$

Substitute $0.100\text{m}/\text{s}$ for $v_{\text{i}}$ in above expression.

    $\begin{array}{c}x=\left(0.100\text{m}/\text{s}\right){\left(0.002293{\left(0.100\text{m}/\text{s}\right)}^2+0.164\right)}^{\frac{1}{2}}+0.04788{\left(0.100\text{m}/\text{s}\right)}^2\\ =0.0410\text{m}\end{array}$

Conclusion:

Therefore, the horizontal position the ball as $v_{\text{i}}=0.100\text{m}/\text{s}$. is $0.0410\text{m}$.

To determine

The horizontal position of the ball with $v_{\text{i}}=100\text{m}/\text{s}$.

Answer to Problem 43AP

The horizontal position the ball with $v_{\text{i}}=100\text{m}/\text{s}$ is $959\text{m}$.

Explanation of Solution

From equation (IV),

    $x=v_{\text{i}}{\left(0.002293v_{\text{i}}{}^2+0.1641\right)}^{\frac{1}{2}}+0.04788v_{\text{i}}{}^2$

Conclusion:

Substitute $100\text{m}/\text{s}$ for $v_{\text{i}}$ in above expression.

    $\begin{array}{c}x=\left(100\text{m}/\text{s}\right){\left(0.002293{\left(100\text{m}/\text{s}\right)}^2+0.1641\right)}^{\frac{1}{2}}+0.04788{\left(100\text{m}/\text{s}\right)}^2\\ =959\text{m}\end{array}$

Therefore, the horizontal position the ball as $v_{\text{i}}=100\text{m}/\text{s}$ is $959\text{m}$.

To determine

The horizontal position of the ball as a function of $v_1$ in a simpler form.

Answer to Problem 43AP

The horizontal position of the ball as a function of $v_{\text{i}}$ in a simpler form is $0.405v_1$.

Explanation of Solution

The located at the spring cannon is $1.2\text{m}$ above the floor and a ball is launched from the cannon with speed $v_{\text{i}}$ at $35°$ above the horizontal.

From equation (IV),

    $x=v_{\text{i}}{\left(0.002293v_{\text{i}}{}^2+0.1641\right)}^{\frac{1}{2}}+0.04788v_{\text{i}}{}^2$

The value of $v_{\text{i}}$ is nearly to zero that is $v_1{}^2\simeq 0$.

Conclusion:

Substitute $0$ for $v_{\text{i}}{}^2$ in above expression.

    $\begin{array}{c}x=v_{\text{i}}{\left(0.002293\times 0+0.16431\right)}^{\frac{1}{2}}+0.04788\times 0\\ =0.405v_{\text{i}}\end{array}$

Therefore, the horizontal position of the ball as a function of $v_{\text{i}}$ in a simpler form is $0.405v_{\text{i}}$.

To determine

The horizontal position of the ball as a function of $v_{\text{i}}$ if the value of $v_{\text{i}}$ is very large.

Answer to Problem 43AP

The horizontal position of the ball as a function of $v_{\text{i}}$ is $0.095v_{\text{i}}{}^2$ if the $v_{\text{i}}$ is very large.

Explanation of Solution

The located at the spring cannon is $1.2\text{m}$ above the floor and a ball is launched from the cannon with speed $v_{\text{i}}$ at $35°$ above the horizontal.

From equation (4),

    $\begin{array}{c}x=v_{\text{i}}{\left(0.002293v_{\text{i}}{}^2+0.1641\right)}^{\frac{1}{2}}+0.04788v_{\text{i}}{}^2\\ ={\left(0.002293v_{\text{i}}{}^4+0.1641v_{\text{i}}{}^2\right)}^{\frac{1}{2}}+0.04788v_{\text{i}}{}^2\end{array}$

Conclusion:

The term is $v_{\text{i}}$ very small compared to the $v_{\text{i}}{}^2$, so neglect the term $0.1643v_{\text{i}}{}^2$.

    $\begin{array}{c}x={\left(0.00229v_{\text{i}}{}^4\right)}^{\frac{1}{2}}+0.0479v_{\text{i}}{}^2\\ =0.0958v_{\text{i}}{}^2\end{array}$

Therefore, the horizontal position of the ball as a function of $v_{\text{i}}$ is $0.095v_{\text{i}}{}^2$ if the $v_{\text{i}}$ is very large.

To determine

The overall shape of the graph of position as a function of velocity.

Answer to Problem 43AP

The starting condition graph $x$ versus $v_{\text{i}}$ is a straight line then with an increase in the value of $v_{\text{i}}$ curve become closer to the parabola.

Explanation of Solution

From the approximation in part (d), it follows that the position curve is a straight line with slope $0.405\text{s}$ for smaller values of $v_{\text{i}}$ and as $v_{\text{i}}$ increases, the curve becomes parabolic following the approximation in part (e).

Conclusion:

Therefore, the starting condition graph $x$ versus $v_{\text{i}}$ is a straight line then with an increase in the value of $v_{\text{i}}$ curve become closer to the parabola.