Chapter 3, Problem 92P

(a)

To determine

The maximum when the ball has reached its first parabolic arc.

Explanation of Solution

Given:

The angle of ball above the horizontal is $55°$ .

The initial speed of the ball is $22\text{m}/\text{s}$ .

The increase in height is $75\%$ .

Formula used:

Write the expression for the horizontal component of the ball.

  $x=x_0+v_{0x}t+\frac{1}{2}{a}_x{t}^2$   ...... (1)

Here, $x_0$ is the initial horizontal component, $v_{0x}$ is the initial velocity of the particle, $a_x$ is the acceleration of the particle and $t$ is the time.

Substitute $0$ for $x_0$ , $v_0\cos {\theta }_0$ for $v_{0x}$ and $0$ for $a_x$ in equation (1).

  $x=v_0\cos {\theta }_{0}t$   ...... (2)

Write the expression for the vertical component of the ball.

  $y=y_0+v_{0y}t+\frac{1}{2}{a}_y{t}^2$   ...... (3)

Here, $y_0$ is the initial vertical component, $v_{0y}$ is the velocity in the vertical direction and $a_y$ is the acceleration in the vertical direction.

Substitute $0$ for $y_0$ , $-g$ for $a_y$ and $v_0\sin {\theta }_0$ for $v_{0y}$ in equation (3).

  $y=v_0\sin {\theta }_{0}t-\frac{1}{2}g{t}^2$   ...... (4)

Write the equation for the

  $v_y=v_{0y}+a_{y}t$

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

  $v_y=v_0\sin {\theta }_0-gt$

When the ball is at the top of trajectory, the vertical velocity at this point is zero.

Substitute $0$ for $v_y$ in the above equation.

  $0=v_0\sin {\theta }_0-gt$

Solve the above equation for $t$ .

  $t=\frac{v_0\sin {\theta }_0}{g}$

Substitute $\frac{v_0\sin {\theta }_0}{g}$ for $t$ and $h$ for $y$ in equation (4).

  $h=v_0\sin {\theta }_0\left(\frac{v_0\sin {\theta }_0}{g}\right)-\frac{1}{2}g{\left(\frac{v_0\sin {\theta }_0}{g}\right)}^2$   ...... (5)

Calculation:

Substitute $55°$ for ${\theta }_0$ , $22\text{m}/\text{s}$ for $v_0$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (5).

  $\begin{array}{l}h=\left(22\text{m}/\text{s}\right)\sin \left(55°\right)\left(\frac{\left(22\text{m}/\text{s}\right)\sin \left(55°\right)}{9.81\text{m}/{\text{s}}^{\text{2}}}\right)-\frac{1}{2}\left(9.81\text{m}/{\text{s}}^{\text{2}}\right){\left(\frac{\left(22\text{m}/\text{s}\right)\sin \left(55°\right)}{9.81\text{m}/{\text{s}}^{\text{2}}}\right)}^2\\ h=16.55\text{m}\end{array}$

Conclusion:

The maximum height reached by the ball at its first parabolic arc is $16.55\text{m}$ .

(b)

To determine

The distance of the ball from the launch point to the ground for the first time.

Explanation of Solution

Given:

The angle of ball above the horizontal is $55°$ .

The initial speed of the ball is $22\text{m}/\text{s}$ .

The increase in height is $75\%$ .

Formula used:

Write the expression for the range of the parabolic path.

  $R_1=\frac{v_0{}^2}{g}\sin 2{\theta }_0$   ...... (6)

Here, $R_1$ is the first range of the ball when it hit the ground.

Calculation:

Substitute $55°$ for ${\theta }_0$ , $22\text{m}/\text{s}$ for $v_0$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (6).

  $\begin{array}{l}{R}_1=\frac{{\left(22\text{m}/\text{s}\right)}^2}{9.81\text{m}/{\text{s}}^{\text{2}}}\sin 2\left(55°\right)\\ R_1=46.36\text{m}\end{array}$

Conclusion:

The distance of the ball from the launch point to the ground for the first time is $46.36\text{m}$ .

(c)

To determine

The distance of the ball from the launch point to the ground for the second time.

Explanation of Solution

Given:

The angle of ball above the horizontal is $55°$ .

The initial speed of the ball is $22\text{m}/\text{s}$ .

The increase in height is $75\%$ .

Formula used:

Write the expression for the total range of the ball.

  $R_T=R_1+R_2$   ...... (7)

Here, $R_T$ is the total range of the ball, $R_1$ is the first range and $R_2$ is the second range when the ball hit the ground.

Write the expression for the range of the parabolic path.

  $R_2=\frac{v_2{}^2}{g}\sin 2{\theta }_0$   ...... (8)

Here, $R_2$ is the first range of the ball when it hit the ground and $v_2$ is the resultant velocity in x and y direction.

Write the expression for the resultant velocity.

  $v_2=\sqrt{v_x^2+v_y^2}$

Here, $v_x$ is the velocity in the x direction and $v_y$ is the velocity in the y direction.

Substitute $v_x^2+v_y^2$ for $v_2{}^2$ in equation (8).

  $R_2=\frac{\left(v_x^2+v_y^2\right)}{g}\sin 2{\theta }_0$   ...... (9)

Write the expression for the relation between initial and final velocity.

  $v_2^2=v_{2y}^2+2a_y{h}^2$

Here, $h$ is the height.

Substitute $0$ for $v_2$ and $-g$ for $h$ in the above equation.

  $\begin{array}{l}0=v_{2y}^2-2g{h}^2\\ v_{2y}^2=2g{h}^2\end{array}$

Substitute $0.75h_1$ for $h_2$ in the above equation.

  $\begin{array}{l}{v}_{2y}^2=2g{\left(0.75h_1\right)}^2\\ v_{2y}^2=1.5g{h}_1{}^2\end{array}$

Write the expression for the rebound angle ${\theta }_2$ .

  ${\theta }_2={\tan }^{-1}\left(\frac{v_{2y}}{v_{2x}}\right)$

Substitute $\sqrt{1.5g{h}_1^2}$ for $v_{2y}$ and $v_0\cos {\theta }_0$ for $v_{2x}$ in the above equation.

  ${\theta }_2={\tan }^{-1}\left(\frac{\sqrt{1.5g{h}_1^2}}{v_0\cos {\theta }_0}\right)$

Substitute $\sqrt{1.5g{h}_1^2}$ for $v_{2y}$ and $v_0\cos {\theta }_0$ for $v_{2x}$ in equation (8).

  $R_2=\frac{1.5g{h}_1^2+v_0^2{\cos }^2{\theta }_0}{g}\sin 2{\theta }_2$   ...... (9)

Write the expression for the

Calculation:

Substitute $22\text{m}/\text{s}$ for $v_0$ , $55°$ for ${\theta }_0$ , $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ , $51.04°$ for ${\theta }_2$ and $16.55\text{m}$ for $h$ .

  $\begin{array}{l}{R}_2=\frac{1.5\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(16.55\text{m}\right)+v_0^2{\cos }^2\left(55°\right)}{9.81\text{m}/{\text{s}}^{\text{2}}}\sin 2\left(51.04°\right)\\ R_2=40.15\text{m}\end{array}$

Substitute $40.15\text{m}$ for $R_2$ and $46.36\text{m}$ for $R_1$ in equation (7).

  $\begin{array}{l}{R}_T=46.36\text{m +}40.15\text{m}\\ R_T=87\text{m}\end{array}$

Conclusion:

The distance of the ball from the launch point to the ground for the second time is $87\text{m}$ .