Chapter 3, Problem 100P

(a)

To determine

The distance where the ball hits the ground.

Explanation of Solution

Given: The distance between the woman and the wall is $4\text{m}$ .

The velocity of the ball is $14\text{m}/\text{s}$ .

The ball is at the distance of $2\text{m}$ above the ground.

The angle made by the ball with the horizontal is $45°$ .

Formula used:

The diagram represents the path followed by the ball.

  

Write the expression for the distance at which the ball hits.

  $x\text{'}=\Delta x-x$

Here $x\text{'}$ is the position where the ball hits the ground, $\Delta x$ is the displacement of the ball after it hits the ground and $x$ is the distance between the person and the wall.

Write the expression for the vertical displacement of the ball.

  $\Delta y=v_{0y}\Delta t-\frac{1}{2}g{\left(\Delta t\right)}^2$   ...... (1)

Here $\Delta y$ is the vertical displacement, $v_{0y}$ is the initial velocity and $\Delta t$ is the time.

The vertical component of velocity is:

  $v_0\sin {\theta }_0$

Substitute $v_0\sin {\theta }_0$ for $v_{0y}$ in equation (1).

  $\Delta y=\left(v_0\sin {\theta }_0\right)\Delta t-\frac{1}{2}g{\left(\Delta t\right)}^2$   ....... (2)

Write the expression for the horizontal distance ball.

  $\Delta x=\left(v_{ox}\cos \theta \right)t$   ...... (3)

Here, $v_{ox}$ is the initial velocity of x direction.

Calculation:

Substitute $45°$ for $\theta$ , $14\text{m}/\text{s}$ for $v_0$ , $-2.0\text{m}$ for $\Delta y$ and $9.18\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (2).

  $-2\text{m}=\left(14\text{m}/\text{s}\right)\left(\sin \left(45°\right)\right)\Delta t-\frac{1}{2}\left(9.18\text{m}/{\text{s}}^{\text{2}}\right){\left(\Delta t\right)}^2$

Solve the above quadratic equation for $\Delta t$ .

  $\Delta t=2.303\text{s}$

Substitute $14\text{m}/\text{s}$ for $v_{ox}$ , $45°$ for $\theta$ and $2.303\text{s}$ for $t$ in equation (3).

  $\begin{array}{l}\Delta x=\left(14\text{m}/\text{s}\right)\cos 45°\left(2.303\text{s}\right)\\ \Delta x=21.8\text{m}\end{array}$

Substitute $21.8\text{m}$ for $\Delta x$ , $4\text{m}$ for $x$ in equation (1).

  $\begin{array}{l}x\text{'}=21.8\text{m}-4\text{m}\\ x\text{'}=18\text{m}\end{array}$

Conclusion:

Thus, the ball hits the ground at $18\text{m}$ .

(b)

To determine

The time for which the ball is in the air.

Explanation of Solution

Given:

The distance between the woman and the wall is $4\text{m}$ .

The velocity of the ball is $14\text{m}/\text{s}$ .

The ball is at the distance of $2\text{m}$ above the ground.

The angle made by the ball with the horizontal is $45°$ .

Formula used:

Write the expression for the time.

  $\Delta t_w=\frac{\Delta x_w}{v_{0x}}$   ...... (4)

Here, $\Delta t_w$ is the time taken by the ball to hit the wall and $\Delta x_w$ is the distance.

Substitute $v_{ox}\cos \theta$ for $v_{ox}$ in equation (4).

  $\Delta t_w=\frac{\Delta x_w}{v_{ox}\cos \theta }$   ...... (5)

Calculation:

Substitute $14\text{m}/\text{s}$

for$45°$ for $\theta$ and $4\text{m}$ for $\Delta x_w$ in equation (5 ).

  $\begin{array}{l}\Delta t_w=\frac{4\text{m}}{\left(14\text{m}/\text{s}\right)\cos \left(45°\right)}\\ \Delta t_w=0.40\text{s}\end{array}$

Conclusion:

Thus, the ball is in the air for $0.40\text{s}$ .

(c)

To determine

The distance at which the ball hits the ground.

Explanation of Solution

Given:

The distance between the woman and the wall is $4\text{m}$ .

The velocity of the ball is

The ball is at the distance of $2\text{m}$above the ground.

The angle made by the ball with the horizontal is

Formula used:

Write the expression for the relation between initial and final velocity in the vertical direction.

  $v_y=v_{0y}+a_y\Delta t$

Substitute $v_{oy}\sin \theta$ for $v_{0y}$ and $-g$ for $a_y$ in the above equation.

  $v_y=v_{oy}\sin \theta -g\Delta t$   ...... (6)

Write the expression to represent the vertical direction.

  $y(t)=y_o+v_{0y}\Delta t+\frac{1}{2}{a}_y{\left(\Delta t\right)}^2$

Substitute $v_0\sin {\theta }_0$ for $v_{0y}$ and $-g$ for $a_y$ in the above equation.

  $y(t)=y_o+\left(v_0\sin {\theta }_0\right)\Delta t-\frac{1}{2}g{\left(\Delta t\right)}^2$   ...... (7)

Calculation:

Substitute $14\text{m}/\text{s}$ for $v_{oy}$ , $45°$ for $\theta$ , $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.40\text{s}$ for $\Delta t$ in equation (6).

  $\begin{array}{l}{v}_y=\left(14\text{m}/\text{s}\right)\sin \left(45°\right)-\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)0.40\text{s}\\ v_y=5.935\text{m}/\text{s}\end{array}$

Substitute $14\text{m}/\text{s}$ for $v_{oy}$ , $45°$ for $\theta$ , $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ , $2\text{m}$ for $y_0$ and $0.40\text{s}$ for $\Delta t$ in equation (7).

  $\begin{array}{l}y(t)=2\text{m}+\left(14\text{m}/\text{s}\left(\sin \left(45°\right)\right)\right)0.40\text{s}-\frac{1}{2}9.81\text{m}/{\text{s}}^{\text{2}}{\left(0.40\text{s}\right)}^2\\ y(t)=5.199\text{m}\end{array}$

Conclusion:

Thus, the distance at which the ball hits the ground is $5.199\text{m}$ .

(d)

To determine

The time for which the ball was in the air after it hit the wall.

Explanation of Solution

Given:

The distance between the woman and the wall is $4\text{m}$ .

The velocity of the ball is

The ball is at the distance of $2\text{m}$ above the ground.

The angle made by the ball with the horizontal is

Formula used:

Write the expression to represent the vertical direction.

  $y(t)=y_o+v_{0y}\Delta t_t+\frac{1}{2}{a}_y{\left(\Delta t_t\right)}^2$

Here, $\Delta t_t$ is the time for which the ball was in the air after it hit the wall.

Substitute $v_0\sin {\theta }_0$ for $v_{0y}$ and $-g$ for $a_y$ in the above equation.

  $y(t)=y_o+\left(v_0\sin {\theta }_0\right)\Delta t_t-\frac{1}{2}g{\left(\Delta t_t\right)}^2$   ...... (8)

When the ball hits the ground, the vertical distance becomes zero.

Calculation:

Substitute $0$ for $y(t)$ , $14\text{m}/\text{s}$ for $v_0$ , $45°$ for $\theta$ , $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $5.199\text{m}$ for $y_0$ equation (8).

  $0=5.199\text{m}+\left(14\text{m}/\text{s}\left(\sin 45°\right)\right)\Delta t_t-\frac{1}{2}9.81\text{m}/{\text{s}}^{\text{2}}{\left(\Delta t_t\right)}^2$

Solve the quadratic equation for $\Delta t_t$ .

  $\Delta t_t=1.798\text{s}$

Conclusion:

Thus, the time for which the ball was in the air after it hit the wall is $1.798\text{s}$ .