Chapter 4, Problem 56AP

(a)

To determine

The total flight time of the ball.

Answer to Problem 56AP

The total fight time of the ball is $2\sqrt{\frac{R}{3g}}$.

Explanation of Solution

The initial velocity is divided into two components that is horizontal and vertical velocity component. At maximum point vertical velocity is zero for this path. The maximum height of the projectile is sixth part of the range.

Write the expression for maximum height of the ball.

    $h_{\max }=\frac{v_i{}^2{\sin }^2{\theta }_i}{2g}$                                                                                            (I)

Here, $h_{\max }$ is maximum height, $v_i$ is initial velocity, $g$ is acceleration due to gravity and ${\theta }_i$ is the angle made by the projectile with the horizontal.

Write the expression for range of the projectile path.

    $R=\frac{v_i{}^2\sin 2{\theta }_i}{g}$

Substitute $2\sin {\theta }_i\cos {\theta }_i$ for $\sin 2{\theta }_i$ in above equation.

    $R=\frac{2v_i{}^2\sin {\theta }_i\cos {\theta }_i}{g}$                                                                                     (II)

Here, $R$ is range of the projectile path.

Multiply equation (II) by $\sin {\theta }_i$ on both sides.

    $R\sin {\theta }_i=\frac{2v_i{}^2{\sin }^2{\theta }_i\cos {\theta }_i}{g}$

Multiply and divide by $2$ in above equation on right side.

    $R\sin {\theta }_i=\frac{4v_i{}^2{\sin }^2{\theta }_i\cos {\theta }_i}{2g}$                                                                         (III)

The maximum height is one-sixth of range.

Substitute $\frac{R}{6}$ for $h_{\max }$ in equation (I)

    $\frac{R}{6}=\frac{v_i{}^2{\sin }^2{\theta }_i}{2g}$

Simplify the above equation.

    $v_i\sin {\theta }_i=\sqrt{\frac{gR}{3}}$                                                                                            (IV)

Substitute $\sqrt{\frac{gR}{3}}$ for $v_i\sin {\theta }_i$ in equation (II).

    $R=\frac{2\left(\sqrt{\frac{gR}{3}}\right)v_i\cos {\theta }_i}{g}$

Simplify and rearrange the above equation as.

    $v_i\cos {\theta }_i=\frac{1}{2}\sqrt{3gR}$                                                                                        (V)

At peak point of the projectile path velocity along vertical direction is zero.

Write the expression for final velocity along vertical.

    $v_{yf}=v_{yi}+a_{y}t$

Here, $v_{yf}$ is the final velocity along vertical, $v_{yi}$ is the initial velocity along vertical, $a_y$ is the vertical acceleration and $t$ is the time.

Initial velocity along vertical is zero.

Substitute $0$ for $v_{yi}$ in above equation.

    $v_{yf}=0+a_{y}t$

Rearrange the above equation.

    $v_{yf}=a_{y}t$

The final velocity along vertical is $v_i\sin {\theta }_i$ and the peak time is $t_{peak}$.

Substitute $v_i\sin {\theta }_i$ for $v_{yf}$  and $t_{peak}$ for $t$ in above equation.

    $v_i\sin {\theta }_i=a_y{t}_{peak}$

Rearrange the above equation time to reach the peak of the path.

  $t_{peak}=\frac{v_i\sin {\theta }_i}{g}$                                                                                                 (VI)

Here, $t_{peak}$ is peak time to reach at maximum height.

Substitute $\sqrt{\frac{gR}{3}}$ for $v_i\sin {\theta }_i$ in equation (VI)

    $t_{peak}=\frac{\left(\sqrt{\frac{gR}{3}}\right)}{g}$

Rearrange the above equation.

    $t_{peak}=\sqrt{\frac{R}{3g}}$

Write the expression for the total time of the ball’s flight.

    $t_{flight}=2t_{peak}$                                                                                             (VII)

Conclusion:

Substitute $\sqrt{\frac{R}{3g}}$ for $t_{peak}$ in equation (VII)

    $t_{flight}=2\sqrt{\frac{R}{3g}}$

Thus, the total flight time of the ball is $2\sqrt{\frac{R}{3g}}$.

(b)

To determine

The speed of the ball at peak point.

Answer to Problem 56AP

The peak velocity at top point is $\frac{1}{2}\sqrt{3gR}$.

Explanation of Solution

At the peak point velocity along vertical is zero and the ball starts to accelerate under gravity in downward direction.

Write the expression for velocity at peak point.

    $v_{peak}=v_i\cos {\theta }_i$                                                                                          (VIII)

Here, $v_{peak}$ is velocity at peak point.

Conclusion:

Substitute $\frac{1}{2}\sqrt{3gR}$ for $v_i\cos {\theta }_i$ in equation (VIII)

    $v_{peak}=\frac{1}{2}\sqrt{3gR}$

Thus, the peak velocity at top point is $\frac{1}{2}\sqrt{3gR}$.

(c)

To determine

The initial vertical component of velocity.

Answer to Problem 56AP

The initial vertical component is $\sqrt{\frac{gR}{3}}$.

Explanation of Solution

Write the expression for initial vertical component of velocity.

    $v_{y_i}=v_i\sin {\theta }_i$                                                                                                (IX)

Here, $v_{y_i}$ is initial vertical component of velocity.

Conclusion:

Substitute $\sqrt{\frac{gR}{3}}$ for $v_i\sin {\theta }_i$ in equation (IX).

    $v_{y_i}=\sqrt{\frac{gR}{3}}$

Thus, the initial vertical component of velocity is $\sqrt{\frac{gR}{3}}$.

(d)

To determine

The initial speed of the ball.

Answer to Problem 56AP

The initial speed of the ball is $\sqrt{\frac{13gR}{12}}$.

Explanation of Solution

The initial speed of the ball is given by the square root of the sum of square of horizontal and vertical velocity component.

Write the expression for initial velocity along horizontal.

    $v_{xi}=v_i\cos {\theta }_i$

Here, $v_{xi}$ is initial velocity along horizontal.

Write the expression for initial velocity along vertical.

    $v_{yi}=v_i\sin {\theta }_i$

Here, $v_{yi}$ is initial velocity along vertical.

Write the expression for the initial speed of ball.

    $v_{(initial)}=\sqrt{v_{xi}^2+v_{yi}^2}$                                                                                   (X)

Here, $v_{(initial)}$ is initial speed of the ball.

Conclusion:

Substitute $v_i\cos {\theta }_i$ for $v_{xi}$ and $v_i\sin {\theta }_i$ for $v_{yi}$ in equation (X).

    $v_{(initial)}=\sqrt{{\left(v_i\cos {\theta }_i\right)}^2+{\left(v_i\sin {\theta }_i\right)}^2}$

Substitute $\frac{1}{2}\sqrt{3gR}$ for $v_i\cos {\theta }_i$ and $\sqrt{\frac{gR}{3}}$ for $v_i\sin {\theta }_i$ in equation (X).

    $v_{(initial)}=\sqrt{{\left(\frac{1}{2}\sqrt{3gR}\right)}^2+{\left(\sqrt{\frac{gR}{3}}\right)}^2}$

Simplify and solve the above equation.

    $v_{(initialspeed)}=\sqrt{\frac{13gR}{12}}$

Thus, the initial speed of the ball is $\sqrt{\frac{13gR}{12}}$.

(e)

To determine

The initial angle of the projectile.

Answer to Problem 56AP

The initial angle of projectile is $33.7°$.

Explanation of Solution

The angle of the projectile can be given by the ratio of vertical velocity to the horizontal velocity.

Write the expression for initial angle of projectile as.

    ${\theta }_i={\tan }^{-1}\left(\frac{v_{yi}}{v_{xi}}\right)$

Substitute $v_i\sin {\theta }_i$ for $v_{yi}$ and $v_i\cos {\theta }_i$ for $v_{xi}$ in above equation.

    ${\theta }_i={\tan }^{-1}\left(\frac{v_i\sin {\theta }_i}{v_i\cos {\theta }_i}\right)$                                                                                   (XI)

Here, ${\theta }_i$ is initial projection angle.

Conclusion:

Substitute $\frac{1}{2}\sqrt{3gR}$ for $v_i\cos {\theta }_i$ and $\sqrt{\frac{gR}{3}}$ for $v_i\sin {\theta }_i$ in equation (XI)

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\sqrt{\frac{gR}{3}}}{\frac{1}{2}\left(\sqrt{3gR}\right)}\right)\\ ={\tan }^{-1}\left(\frac{2}{3}\right)\\ =33.7\end{array}$

Thus, the initial projection angle is $33.7°$.

(f)

To determine

The greatest height with same initial speed.

Answer to Problem 56AP

The greatest height at same initial speed is $\frac{13}{24}R$.

Explanation of Solution

The maximum height can be achieved by the ball at $90°$ in the projectile motion.

Conclusion:

Substitute $90°$ for $\theta$ and $\sqrt{\frac{13gR}{12}}$ for $v_i$ in equation (I)

    $\begin{array}{c}{h}_{\max }=\frac{{\left(\sqrt{\frac{13gR}{12}}\right)}^2\sin 90°}{2g}\\ =\frac{13gR}{12\left(2g\right)}\\ =\frac{13}{24}R\end{array}$

Thus, the greatest height at same initial speed is $\frac{13}{24}R$.

(g)

To determine

The maximum horizontal range.

Answer to Problem 56AP

The maximum horizontal range is $\frac{13}{12}R$.

Explanation of Solution

The maximum horizontal range can be achieved at $45°$ in the projectile motion.

Write the expression for maximum range as.

    $R_{\max }=\frac{v_i^2\sin 2{\theta }_i}{g}$                                                                                        (XII)

Here, $R_{\max }$ is the maximum range.

Conclusion:

Substitute $45°$ for $\theta$ and $\sqrt{\frac{13gR}{12}}$ for $v_i$ in equation (XII)

    $\begin{array}{c}{R}_{\max }=\frac{{\left(\sqrt{\frac{13gR}{12}}\right)}^2\sin 2\left(45°\right)}{g}\\ =\frac{13gR\left(\sin 90°\right)}{12g}\\ =\frac{13}{12}R\end{array}$

Thus, the maximum horizontal range is $\frac{13}{12}R$.