Chapter 4, Problem 4.56AP

A ball is thrown with an initial speed υ_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

The time interval during which the ball is in motion.

Answer to Problem 4.56AP

The time interval during which the ball is in motion is $\sqrt{\frac{4R}{3g}}$.

Explanation of Solution

Given info: The initial speed of the ball is $v_i$ and the angle made by the ball with the horizontal is ${\theta }_{\text{i}}$, the horizontal range of the ball is $R$ and the maximum height covered by the ball is $\frac{R}{6}$

The motion of the ball follows the parabolic path and the ball is said to projectile, the motion of the ball is shown in the Figure below.

Figure (1)

The formula to calculate the maximum height reached by the projectile is,

$H=\frac{{\left(v_i\right)}^2{\sin }^2\theta }{2g}$

Here,

$g$ is the acceleration due to gravity.

$H$ is the maximum height reached by the ball.

Rearrange the above equation.

$\begin{array}{c}{\left(v_i\right)}^2{\sin }^2\theta =2gH\\ v_i\sin \theta =\sqrt{2gH}\end{array}$

Substitute $\frac{R}{6}$ for $H$ in the above equation.

$\begin{array}{c}{v}_i\sin \theta =\sqrt{\frac{2gR}{6}}\\ =\sqrt{\frac{gR}{3}}\end{array}$

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

The formula to calculate the time taken by the ball to reach the ground is,

$t=\frac{2v_{\text{i}}\sin \theta }{g}$

Here,

$t$ is the time taken by the ball to reach the ground.

Substitute $\sqrt{\frac{gR}{3}}$ for $v_i\sin \theta$ in the above equation.

$\begin{array}{c}t=\frac{2\sqrt{\frac{gR}{3}}}{g}\\ =\sqrt{\frac{4R}{3g}}\end{array}$

Conclusion:

Therefore, the time interval during which the ball is in motion is $\sqrt{\frac{4R}{3g}}$.

To determine

The speed of the ball at the peak of its path.

Answer to Problem 4.56AP

The speed of the ball at the peak of its path is $\sqrt{\frac{3gR}{4}}$.

Explanation of Solution

Given info: The initial speed of the ball is $v_i$ and the angle made by the ball with the horizontal is ${\theta }_{\text{i}}$, the horizontal range of the ball is $R$ and the maximum height covered by the ball is $\frac{R}{6}$

From part (a) the time of flight is $\sqrt{\frac{4R}{3g}}$.

From the Figure (1) the range of the ball and time of flight is,

$R=\left(v_{\text{i}}\cos {\theta }_{\text{i}}\right)t$

Rearrange the above equation.

$v_{\text{i}}\cos {\theta }_{\text{i}}=\frac{R}{t}$

Substitute $\sqrt{\frac{4R}{3g}}$ for $t$ in the above equation.

$\begin{array}{c}{v}_{\text{i}}\cos {\theta }_{\text{i}}=\frac{R}{\sqrt{\frac{4R}{3g}}}\\ =\sqrt{\frac{3g{R}^2}{4R}}\\ =\sqrt{\frac{3gR}{4}}\end{array}$

Conclusion:

Therefore, the speed of the ball at the peak of its path is $\sqrt{\frac{3gR}{4}}$.

To determine

The initial vertical component of the velocity.

Answer to Problem 4.56AP

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

Explanation of Solution

Given info: The initial speed of the ball is $v_i$ and the angle made by the ball with the horizontal is ${\theta }_{\text{i}}$, the horizontal range of the ball is $R$ and the maximum height covered by the ball is $\frac{R}{6}$

From part (a) vertical component of the initial velocity is $\sqrt{\frac{gR}{3}}$.

Conclusion:

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

To determine

The initial speed of the ball.

Answer to Problem 4.56AP

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

Explanation of Solution

Given info: The initial speed of the ball is $v_i$ and the angle made by the ball with the horizontal is ${\theta }_i$, the horizontal range of the ball is $R$ and the maximum height covered by the ball is $\frac{R}{6}$

From part (a) vertical component of the initial velocity is,

$v_{\text{i}}\sin {\theta }_{\text{i}}=\sqrt{\frac{gR}{3}}$

Square both side of the above equation.

$\begin{array}{c}{\left(v_i\sin {\theta }_i\right)}^2={\left(\sqrt{\frac{gR}{3}}\right)}^2\\ =\frac{gR}{3}\end{array}$ (1)

And from part (b) the horizontal component of the velocity is $\sqrt{\frac{3gR}{4}}$

$v_{\text{i}}\cos {\theta }_{\text{i}}=\sqrt{\frac{3gR}{4}}$

Square both side of the above equation.

$\begin{array}{c}{\left(v_{\text{i}}\cos {\theta }_{\text{i}}\right)}^2={\left(\sqrt{\frac{3gR}{4}}\right)}^2\\ =\frac{3gR}{4}\end{array}$ (2)

Add equation (1) and (2) to find the initial velocity.

$\begin{array}{c}{\left(v_{\text{i}}\sin {\theta }_{\text{i}}\right)}^2+{\left(v_{\text{i}}\cos {\theta }_{\text{i}}\right)}^2=\frac{3gR}{4}+\frac{gR}{3}\\ {\left(v_{\text{i}}\right)}^2\left({\left(\sin {\theta }_{\text{i}}\right)}^2+{\left(\cos {\theta }_{\text{i}}\right)}^2\right)=\frac{9gR+4gR}{12}\\ {\left(v_{\text{i}}\right)}^2=\frac{13gR}{12}\\ v_{\text{i}}=\sqrt{\frac{13gR}{12}}\end{array}$

Conclusion:

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

To determine

The angle ${\theta }_{\text{i}}$.

Answer to Problem 4.56AP

The angle ${\theta }_{\text{i}}$ is $33.7°$.

Explanation of Solution

Given info: The initial speed of the ball is $v_i$ and the angle made by the ball with the horizontal is ${\theta }_{\text{i}}$, the horizontal range of the ball is $R$ and the maximum height covered by the ball is $\frac{R}{6}$

From part (a) vertical component of the initial velocity is,

$v_{\text{i}}\sin {\theta }_{\text{i}}=\sqrt{\frac{gR}{3}}$

And from part (b) the horizontal component of the velocity is $\sqrt{\frac{3gR}{4}}$

$v_{\text{i}}\cos {\theta }_{\text{i}}=\sqrt{\frac{3gR}{4}}$

Take the ratio of the horizontal component and the vertical component of the initial velocity.

$\begin{array}{c}\frac{v_{\text{i}}\sin {\theta }_{\text{i}}}{v_{\text{i}}\cos {\theta }_{\text{i}}}=\frac{\sqrt{\frac{gR}{3}}}{\sqrt{\frac{3gR}{4}}}\\ \tan {\theta }_{\text{i}}=\sqrt{\frac{4gR}{9gR}}\\ {\theta }_{\text{i}}={\tan }^{-1}\sqrt{\frac{9}{4}}\\ =33.7°\end{array}$

Conclusion:

Therefore, the angle ${\theta }_{\text{i}}$ is $33.7°$.

To determine

The maximum height that the ball can reach with the same initial velocity.

Answer to Problem 4.56AP

The maximum height that the ball can reach with the same initial velocity is       $\frac{13R}{24}$.

Explanation of Solution

Given info: The initial speed of the ball is $v_i$ and the angle made by the ball with the horizontal is ${\theta }_{\text{i}}$, the horizontal range of the ball is $R$ and the maximum height covered by the ball is $\frac{R}{6}$

For the maximum height to be gained by the ball the angle made by the horizontal should be $90°$.

The formula to calculate the maximum height reached by the projectile is,

$H_{\max }=\frac{{\left(v_{\text{i}}\right)}^2{\sin }^{2}90°}{2g}$

Here,

$H_{\max }$ is the maximum height reached by the ball.

Rearrange the above equation.

$\begin{array}{c}{H}_{\max }=\frac{{\left(v_{\text{i}}\right)}^2{\sin }^{2}90°}{2g}\\ =\frac{{\left(v_{\text{i}}\right)}^2}{2g}\end{array}$

From part (d) the initial velocity of the ball is,

$v_{\text{i}}=\sqrt{\frac{13gR}{12}}$

Substitute $\sqrt{\frac{13gR}{12}}$ for $v_i$ in the above equation.

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

Conclusion:

Therefore, the maximum height that the ball can reach with the same initial velocity is       $\frac{13R}{24}$.

To determine

The maximum range of the ball with the same initial velocity.

Answer to Problem 4.56AP

The maximum range of the ball with the same initial velocity is $\frac{13R}{12}$.

Explanation of Solution

Given info: The initial speed of the ball is $v_i$ and the angle made by the ball with the horizontal is ${\theta }_{\text{i}}$, the horizontal range of the ball is $R$ and the maximum height covered by the ball is $\frac{R}{6}$

For the maximum range to be gained by the ball the angle made by the horizontal should be $45°$.

The formula to calculate the maximum height reached by the projectile is,

$\begin{array}{c}{R}_{\max }=\frac{{\left(v_{\text{i}}\right)}^2{\sin }^{2}2\left(45°\right)}{g}\\ =\frac{{\left(v_{\text{i}}\right)}^2}{g}\end{array}$

Here,

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

From part (d) the initial velocity of the ball is,

$v_{\text{i}}=\sqrt{\frac{13gR}{12}}$

Substitute $\sqrt{\frac{13gR}{12}}$ for $v_i$ in the above equation.

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

Conclusion:

Therefore, the maximum range of the ball with the same initial velocity is $\frac{13R}{12}$.