Chapter 3, Problem 94P

(a)

To determine

The distance of the ball strikes the wall.

Answer to Problem 94P

The distance of the ball strikes the wall is $1.1\text{m}$.

Explanation of Solution

Write the equation of motion for the ball taken to reach the wall.

$\Delta x=v_x\Delta t$ (I)

Here, $\Delta x$ is the displacement of the ball, $v_x$ is the velocity of the ball along x axis, and $\Delta t$ is the time taken by the ball to reach the wall.

Write the equation of motion for the ball’s initial velocity along x axis.

$v_x=v_{\text{i}}\cos \theta$ (II)

Here, $v_{\text{i}}$ is the initial velocity of the ball and $\theta$ is the vertical angle above the ground.

Write the equation of motion for the ball’s initial velocity along y axis.

$v_{\text{i}y}=v_{\text{i}}\sin \theta$ (III)

Here, $v_{\text{i}}$ is the initial velocity of the ball and $\theta$ is the vertical angle above the ground.

Write the equation of motion for the ball reaches certain height.

$\Delta y=v_{\text{i}y}\Delta t+\frac{1}{2}{a}_y\Delta t^2$

Here, $\Delta y$ is the change in height of the ball, $v_{\text{i}y}$ is the initial velocity of the ball along y axis, and $a_y$ is the acceleration of the ball.

Rewrite the above equation for gravitational acceleration.

$\Delta y=v_{\text{i}y}\Delta t-\frac{1}{2}g\Delta t^2$ (IV)

Here, $g$ is the gravitational acceleration.

Write the equation for change in height of the ball.

$\Delta y=y_{\text{f}}-y_{\text{i}}$ (V)

Here, $y_{\text{f}}$ is the final height and $y_{\text{i}}$ is the initial height.

Conclusion:

Substitute equation (II) in equation (I) to find $\Delta t$.

$\begin{array}{c}\Delta x=\left(v_{\text{i}}\cos \theta \right)\Delta t\\ \Delta t=\frac{\Delta x}{v_{\text{i}}\cos \theta }\end{array}$ (VI)

Substitute equation (III) and (VI) in equation (IV) to find $\Delta y$.

$\begin{array}{c}\Delta y=\left(v_{\text{i}}\sin \theta \right)\Delta t-\frac{1}{2}g\Delta t^2\\ =\left(v_{\text{i}}\sin \theta \right)\frac{\Delta x}{v_{\text{i}}\cos \theta }-\frac{1}{2}g\left(\frac{\Delta x^2}{v_{\text{i}}^2{\cos }^2\theta }\right)\\ =\Delta x\tan \theta -\left(\frac{g\Delta x^2}{2v_{\text{i}}^2{\cos }^2\theta }\right)\end{array}$

Substitute the above relation in equation (V) to find $y_{\text{f}}$.

$\begin{array}{c}{y}_{\text{f}}=y_{\text{i}}+\Delta y\\ =y_{\text{i}}+\Delta x\tan \theta -\left(\frac{g\Delta x^2}{2v_{\text{i}}^2{\cos }^2\theta }\right)\end{array}$

Substitute $60\text{cm}$ for $y_{\text{i}}$, $10\text{m}$ for $\Delta x$, $9.80{\text{m/s}}^2$ for $g$, $20\text{m/s}$ for $v_{\text{i}}$, and $90°-80°=10°$ for $\theta$ in the above equation.

$\begin{array}{c}{y}_{\text{f}}=60\text{cm}\left(\frac{0.01\text{m}}{1\text{cm}}\right)+\left(10\text{m}\right)\tan 10°-\left(\frac{\left(9.80{\text{m/s}}^2\right){\left(10\text{m}\right)}^2}{2{\left(20\text{m/s}\right)}^2{\cos }^{2}10°}\right)\\ =2.36\text{m}-\frac{980{\text{m}}^3/{\text{s}}^2}{775.9{\text{m}}^2/{\text{s}}^2}\\ =2.36\text{m}-1.26\text{m}\\ =1.1\text{m}\end{array}$

Therefore, the distance of the ball strikes the wall is $1.1\text{m}$.

(b)

To determine

Whether the ball rolls up or down when it hits the wall.

Answer to Problem 94P

The ball rolls down when it hits the wall.

Explanation of Solution

Write the equation for change in velocity of the ball along y axis.

$\Delta v=v_{\text{fy}}-v_{\text{iy}}$ (VII)

Here, $v_{\text{fy}}$ is the final velocity and $v_{\text{iy}}$ is the initial velocity.

Write the equation of motion for the acceleration of ball.

$\Delta v=a_y\Delta t$ (VIII)

Here, $\Delta v$ is the change in velocity.

Rewrite the equation (VIII) for gravitational acceleration.

$\Delta v=-g\Delta t$ (IX)

Conclusion:

Compare equation (VII) and (IX).

$v_{\text{fy}}-v_{\text{iy}}=-g\Delta t$

Replace $v_{\text{i}}\sin \theta$ for $v_{\text{i}y}$ and $\frac{\Delta x}{v_{\text{i}}\cos \theta }$ for $\Delta t$ in above relation.

$\begin{array}{c}{v}_{\text{fy}}-v_{\text{i}}\sin \theta =-g\left(\frac{\Delta x}{v_{\text{i}}\cos \theta }\right)\\ v_{\text{fy}}=v_{\text{i}}\sin \theta -\left(\frac{g\Delta x}{v_{\text{i}}\cos \theta }\right)\end{array}$ (X)

Substitute $10\text{m}$ for $\Delta x$, $9.80{\text{m/s}}^2$ for $g$, $20\text{m/s}$ for $v_{\text{i}}$, and $90°-80°=10°$ for $\theta$ in equation (X).

$\begin{array}{c}{v}_{\text{fy}}=\left(20\text{m/s}\right)\sin 10°-\left(\frac{\left(9.80{\text{m/s}}^2\right)\left(10\text{m}\right)}{\left(20\text{m/s}\right)\cos 10°}\right)\\ =3.47\text{m/s}-4.98\text{m/s}\\ =-1.51\text{m/s}\\ \simeq -\text{1}\text{.5}\text{m/s}\end{array}$

Since,$v_{\text{fy}}<0$, so that ball is on its way down.