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 horizontal position of the ball as a function of $v_1$.

Answer to Problem 43AP

The horizontal position of the ball as a function of $v_1$ is $v_1{\left(0.00229v_1{}^2+0.1643\right)}^{\frac{1}{2}}+0.0479v_1{}^2$.

Explanation of Solution

Given info: The located at the spring cannon is $1.2\text{m}$ above the floor and a ball is launched from the cannon with speed $v_1$ at $35°$ above the horizontal.

Formula to calculate the vertical distance covered by the ball is,

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

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.

$y_{\text{o}}$ is the initial vertical distance above the floor.

The vertical component of the velocity is,

$v_{\text{y}}=v_1\sin \theta$

Here,

$v_1$ is the velocity of the ball during projectile motion.

Substitute $v_1\sin \theta$ for $v_{\text{y}}$ in equation (1).

$y=y_{\text{o}}+v_1\sin \theta \times t-\frac{1}{2}g{t}^2$

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 $t$.

$\begin{array}{c}0=1.2\text{m}+v_1\sin 35.0°\times t-\frac{1}{2}9.81\text{m}/{\text{s}}^{\text{2}}\times t^2\\ 0=9.81\text{m}/{\text{s}}^{\text{2}}\times t^2-2v_1\sin 35.0°\times t-1.2\text{m}\end{array}$ (2)

Solve the equation (2).

$t=0.05846v_1+{\left(3.4184\times {10}^{-3}{v}_1{}^2+0.2446\right)}^{\frac{1}{2}}$

Formula to calculate the horizontal distance covered by the ball is,

$x=v_{\text{x}}t$ (3)

Here,

$v_{\text{x}}$ is the component of velocity in x direction.

$t$ is the time interval.

The horizontal component of the velocity is,

$v_{\text{x}}=v_1\cos \theta$

Substitute $v_1\cos \theta$ for $v_{\text{x}}$ and $0.05846v_1+{\left(3.4184\times {10}^{-3}{v}_1{}^2+0.2446\right)}^{\frac{1}{2}}$ for $t$ in equation (3).

$x=v_1\cos \theta \times \left(0.05846v_1+{\left(3.4184\times {10}^{-3}{v}_1{}^2+0.2446\right)}^{\frac{1}{2}}\right)$

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

$\begin{array}{c}x=v_1\cos 35.0°\times \left(0.05846v_1+{\left(3.4184\times {10}^{-3}{v}_1{}^2+0.2446\right)}^{\frac{1}{2}}\right)\\ =v_1{\left(0.00229v_1{}^2+0.1643\right)}^{\frac{1}{2}}+0.0479v_1{}^2\end{array}$ (4)

Conclusion:

Therefore, the horizontal position of the ball as a function of $v_1$ is $v_1{\left(0.00229v_1{}^2+0.1643\right)}^{\frac{1}{2}}+0.0479v_1{}^2$.

To determine

The horizontal position of the ball as $v_1=0.1\text{m}/\text{s}$.

Answer to Problem 43AP

The horizontal position the ball as $v_1=0.1\text{m}/\text{s}$. is $0.041\text{m}$.

Explanation of Solution

Given info: The located at the spring cannon is $1.2\text{m}$ above the floor and a ball is launched from the cannon with speed $v_1$ at $35°$ above the horizontal.

From equation (IV),

$x=v_1{\left(0.00229v_1{}^2+0.1643\right)}^{\frac{1}{2}}+0.0479v_1{}^2$

Substitute $0.1\text{m}/\text{s}$ for $v_1$ in above expression.

$\begin{array}{c}x=0.1\text{m}/\text{s}{\left(0.00229{\left(0.1\text{m}/\text{s}\right)}^2+0.1643\right)}^{\frac{1}{2}}+0.0479{\left(0.1\text{m}/\text{s}\right)}^2\\ =0.041\text{m}\end{array}$

Conclusion:

Therefore, the horizontal position the ball as $v_1=0.1\text{m}/\text{s}$. is $0.041\text{m}$.

To determine

The horizontal position of the ball as $v_1=100\text{m}/\text{s}$.

Answer to Problem 43AP

The horizontal position the ball as $v_1=100\text{m}/\text{s}$. is $960.18\text{m}$.

Explanation of Solution

Given info: The located at the spring cannon is $1.2\text{m}$ above the floor and a ball is launched from the cannon with speed $v_1$ at $35°$ above the horizontal.

From equation (4),

$x=v_1{\left(0.00229v_1{}^2+0.1643\right)}^{\frac{1}{2}}+0.0479v_1{}^2$

Substitute $100\text{m}/\text{s}$ for $v_1$ in above expression.

$\begin{array}{c}x=100\text{m}/\text{s}{\left(0.00229{\left(100\text{m}/\text{s}\right)}^2+0.1643\right)}^{\frac{1}{2}}+0.0479{\left(100\text{m}/\text{s}\right)}^2\\ =960.18\text{m}\end{array}$

Conclusion:

Therefore, the horizontal position the ball as $v_1=0.1\text{m}/\text{s}$. is $960.18\text{m}$.

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 $v_1$ in a simpler form is $0.405v_1$.

Explanation of Solution

Given info: The located at the spring cannon is $1.2\text{m}$ above the floor and a ball is launched from the cannon with speed $v_1$ at $35°$ above the horizontal.

From equation (IV),

$x=v_1{\left(0.00229v_1{}^2+0.1643\right)}^{\frac{1}{2}}+0.0479v_1{}^2$

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

Substitute $0$ for $v_1{}^2$ in above expression.

$\begin{array}{c}x=v_1{\left(0.00229\times 0+0.1643\right)}^{\frac{1}{2}}+0.0479\times 0\\ =0.405v_1\end{array}$

Conclusion:

Therefore, the horizontal position of the ball as a function of $v_1$ in a simpler form is $0.405v_1$.

To determine

The horizontal position of the ball as a function of $v_1$ if the value of $v_1$ is very large.

Answer to Problem 43AP

The horizontal position of the ball as a function of $v_1$ is $0.095v_1{}^2$ if the $v_1$ is very large.

Explanation of Solution

Given info: The located at the spring cannon is $1.2\text{m}$ above the floor and a ball is launched from the cannon with speed $v_1$ at $35°$ above the horizontal.

From equation (4),

$\begin{array}{c}x=v_1{\left(0.00229v_1{}^2+0.1643\right)}^{\frac{1}{2}}+0.0479v_1{}^2\\ ={\left(0.00229v_1{}^4+0.1643v_1{}^2\right)}^{\frac{1}{2}}+0.0479v_1{}^2\end{array}$

The term is $0.1643v_1{}^2$ very small compare to the $0.00229v_1{}^4$ so the neglect the term $0.1643v_1{}^2$.

$\begin{array}{c}x={\left(0.00229v_1{}^4\right)}^{\frac{1}{2}}+0.0479v_1{}^2\\ =0.095v_1{}^2\end{array}$

Conclusion:

Therefore, the horizontal position of the ball as a function of $v_1$ is $0.095v_1{}^2$ if the $v_1$ is very large.

To determine

The overall shape of the graph of the function $x\left(v\right)$.

Answer to Problem 43AP

In starting condition graph $x$ versus $v_1$ is a straight line then increase the value of $v_1$ curve become closer to the parabola.

Explanation of Solution

Given info: The located at the spring cannon is $1.2\text{m}$ above the floor and a ball is launched from the cannon with speed $v_1$ at $35°$ above the horizontal.

The graph of $x$ versus $v_1$ start from the origin as a straight line with the slope is $0.405\text{s}$ then it curves upward above this tangent line, becoming a closer and closer to the parabola.

Conclusion:

Therefore, the starting condition graph $x$ versus $v_1$ is a straight line then increase the value of $v_1$ curve become closer to the parabola.