Chapter 3, Problem 54P

A ball is thrown with an initial speed v_i at an angle θ_i with the horizontal. The horizontal range of the ball is R, and the ball reaches a maximum height R/6. In terms of R and g, find (a) the time interval during which the ball is in motion, (b) the ball’s speed at the peak of its path, (c) the initial vertical component of its velocity, (d) its initial speed, and (e) the angle θ_i. (f) Suppose the ball is thrown at the same initial speed found in (d) but at the angle appropriate for reaching the greatest height that it can. Find this height. (g) Suppose the ball is thrown at the same initial speed but at the angle for greatest possible range. Find this maximum horizontal range.

To determine

Theflight time of the ball in the motion .

Answer to Problem 54P

The flight time of the ball in the motion is $\underset{\_}{2\sqrt{\frac{R}{3g}}}$.

Explanation of Solution

Write the expression for the maximum height of the ball,

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

Here, $h$ is the maximum height of the ball, $v_i$ is the initial speed of the ball, ${\theta }_i$ is the angle and $g$ is the acceleration due to gravity.

Write the expression for the horizontal range of the ball,

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

Here, $R$ is the horizontal range of the ball.

Substitute $\frac{R}{6}$ for $h$ in (I),

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

Combine (I) and (II) and substitute $\frac{R}{6}$ for $h$ and $\sqrt{\frac{gR}{3}}$ for $v_i\sin {\theta }_i$,

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

Rewrite the relation for $v_i$.

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

Write the expression for the vertical velocity of the ball,

    $v_{yf}=v_{yi}+a_{y}t$        (VI)

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

Conclusion:

Substitute $0$ for $v_{yf}$, $v_i\sin {\theta }_i$ for $v_{yi}$, $t_{peak}$ for $t$ and $g$ for $a_y$ in (VI),

  $\begin{array}{c}0=v_i\sin {\theta }_i+g{t}_{peak}\\ t_{peak}=v_i\sin {\theta }_i/g\\ t_{peak}=\sqrt{\frac{gR}{3}}/g\\ =\sqrt{\frac{R}{3g}}\end{array}$

Total time of the ball’s flight,

    $\begin{array}{c}{t}_{flight}=2t_{peak}\\ =2\sqrt{\frac{R}{3g}}\end{array}$

Therefore, theflight time of the ball in the motion is $\underset{\_}{2\sqrt{\frac{R}{3g}}}$.

To determine

Thespeed of the ball at the peak of its path.

Answer to Problem 54P

The speed of the ball at the peak of its path is $\underset{\_}{v_{xi}=\frac{1}{2}\sqrt{3gR}}$.

Explanation of Solution

Write the expression for the speed of the ball at the path’s peak,

    $\begin{array}{c}{v}_{peak}=v_{xi}\\ =v_i\cos {\theta }_i\end{array}$        (VII)

Here, $v_{peak}$ is the speed of the ball at the path’s peak, $v_{xi}$ is the initial horizontal velocity of the ball, $v_i$ is the initial speed of the ball and ${\theta }_i$ is the projectile angle of the ball.

Conclusion:

Substitute $\frac{1}{2}\sqrt{3gR}$ for $v_i\cos {\theta }_i$ in the above equation,

    $v_{peak}=\frac{1}{2}\sqrt{3gR}$

Therefore, thespeed of the ball at the peak of its path is $\underset{\_}{v_{xi}=\frac{1}{2}\sqrt{3gR}}$.

To determine

The initial vertical component of the ball’s velocity .

Answer to Problem 54P

The initial vertical component of the ball’s velocity is $\underset{\_}{v_{yi}=\sqrt{\frac{gR}{3}}}$.

Explanation of Solution

Write the expression for the initial vertical velocity of the ball,

    $v_{yi}=v_i\sin {\theta }_i$        (VIII)

Here, $v_{yi}$ is the initial vertical velocity of the ball, $v_i$ is the initial speed of the ball and ${\theta }_i$ is the projectile angle of the ball.

Conclusion:

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

    $v_{yi}=\sqrt{\frac{gR}{3}}$

Therefore, theinitial vertical component of the ball’s velocity is $\underset{\_}{v_{yi}=\sqrt{\frac{gR}{3}}}$.

To determine

The initial speed of the ball.

Answer to Problem 54P

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

Explanation of Solution

Write the expression for the initial speed of the ball,

    $v_i^2=v_{xi}^2+v_{yi}^2$        (IX)

Here, $v_i$ is the initial speed of the ball, $v_{yi}$ is the initial vertical velocity of the ball, $v_{xi}$ is the initial horizontal velocity of the ball and ${\theta }_i$ is the projectile angle of the ball.

Rewrite the above expression,

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

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 (X),

    $\begin{array}{c}{v}_i^2={\left(\sqrt{\frac{gR}{3}}\right)}^2+{\left(\frac{1}{2}\sqrt{3gR}\right)}^2\\ =\frac{gR}{3}+\frac{3gR}{4}\\ =\frac{13gR}{12}\end{array}$

Therefore, theinitial speed of the ball is $\underset{\_}{v_i=\sqrt{\frac{13gR}{12}}}$.

To determine

The projectile angle of the ball .

Answer to Problem 54P

The projectile angle of the ballis $\underset{\_}{{\theta }_i={33.7}^{\text{o}}}$.

Explanation of Solution

Equation (III) divided by (V),

    $\begin{array}{l}\frac{v_i\sin {\theta }_i}{v_i\cos {\theta }_i}=\left[\frac{\sqrt{gR/3}}{\left(\frac{1}{2}\sqrt{3gR}\right)}\right]\\ \tan {\theta }_i=\frac{2}{3}\end{array}$

Conclusion:

Rewrite the above equation,

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

Therefore, theprojectile angle of the ball is $\underset{\_}{{\theta }_i={33.7}^{\text{o}}}$.

To determine

The maximum height of the ball when it throws at maximum projection angle.

Answer to Problem 54P

The maximum height of the ball when it throws at maximum projection angle is $\underset{\_}{\frac{13}{24}R}$.

Explanation of Solution

In this case, the maximum projection angle must be ${90.0}^{\text{o}}$ .

From (I),

Write the expression for the maximum height of the ball thrown,

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

Here, $h_{\max }$ is the maximum height of the ball thrown.

Conclusion:

Substitute $\frac{13}{12}gR$ for $v_i$ and ${90.0}^{\text{o}}$ for ${\theta }_i$ in the above equation,

    $\begin{array}{c}{h}_{\max }=\frac{\left(\frac{13}{12}gR\right){\sin }^2{90.0}^{\text{o}}}{2g}\\ =\frac{13}{24}R\end{array}$

Therefore, themaximum height of the ball when it throws at maximum projection angle is $\underset{\_}{\frac{13}{24}R}$.

To determine

The maximum horizontal range of the ball.

Answer to Problem 54P

The maximum horizontal range of the ballis $\underset{\_}{\frac{13}{12}R}$.

Explanation of Solution

In this case, the maximum projection angle must be ${90.0}^{\text{o}}$ .

Write the expression for the maximum range of the ball thrown,

    $R_{\max }=\frac{v_i\sin {\theta }_i}{g}$

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

Conclusion:

Substitute $\frac{13}{12}gR$ for $v_i$ and ${90.0}^{\text{o}}$ for ${\theta }_i$ in the above equation,

    $\begin{array}{c}{R}_{\max }=\frac{\left(\frac{13}{12}gR\right){\sin }^2{90.0}^{\text{o}}}{g}\\ =\frac{13}{12}R\end{array}$

Therefore, themaximum horizontal range of the ball is $\underset{\_}{\frac{13}{12}R}$.