Chapter 6, Problem 67CP

A golfer tees off from a location precisely at ϕ_i = 35.0° north latitude. He hits the ball due south, with range 285 m. The ball’s initial velocity is at 48.0° above the horizontal. Suppose air resistance is negligible for the golf ball. (a) For how long is the ball in flight? The cup is due south of the golfer’s location, and the golfer would have a hole-in-one if the Earth were not rotating. The Earth’s rotation makes the tee move in a circle of radius R_E cos ϕ_i = (6.37 × 10⁶ m) cos 35.0° as shown in Figure P6.47. The tee completes one revolution each day. (b) Find the eastward speed of the tee relative to the stars. The hole is also moving cast, but it is 285 m farther south and thus at a slightly lower latitude ϕ_f. Because the hole moves in a slightly larger circle, its speed must he greater than that of the tee. (c) By how much does the hole’s speed exceed that of the tee? During the time interval the ball is in flight, it moves upward and downward as well as southward with the projectile motion you studied in Chapter 4, but it also moves eastward with the speed you found in part (b). The hole moves to the east at a faster speed, however, pulling ahead of the ball with the relative speed you found in part (c). (d) How far to the west of the hole does the ball land?

Figure P6.47

To determine

The time for which the ball be in flight.

Answer to Problem 67CP

The time for which the ball be in flight is $8.04\sec$.

Explanation of Solution

Given information: The range of the motion after hitting the ball is $285\text{m}$. The direction of initial velocity of ball is $48.0°$ above the horizontal, the radius of the circular path is $R_{\text{E}}\cos {\varphi }_{\text{i}}=\left(6.37\times {10}^6\text{m}\right)\cos 35.0°$ and the location from which the golfer tees off is $35.0°$ north latitude.

The range of the parabolic motion is given as,

$R=V_0\cos \beta t$ (I)

• $V_0$ is the initial velocity.
• $\beta$ is the direction of initial velocity.
• $t$ is the time.

The equation for parabolic motion is given as,'

$\begin{array}{l}y=y_0+V_0\sin \beta t-\frac{g{t}^2}{2}\\ y-y_0=V_0\sin \beta t-\frac{g{t}^2}{2}\end{array}$

As initial and final distance is equal, $y-y_0=0$.

$\begin{array}{l}{V}_0\sin \beta t-\frac{g{t}^2}{2}=0\\ V_0\sin \beta t=\frac{g{t}^2}{2}\end{array}$

Rearrange the above expression for $t$.

$t=\frac{2V_0\sin \beta }{g}$ (II)

Substitute $\frac{2V_0\sin \beta }{g}$ for $t$ in equation (I).

$\begin{array}{c}R=V_0\cos \beta \left(\frac{2V_0\sin \beta }{g}\right)\\ R=\frac{2V_0^2\cos \beta \sin \beta }{g}\\ R=\frac{V_0^2\sin 2\beta }{g}\end{array}$

Rearrange the above expression for $V_0$.

$V_0=\sqrt{\frac{gR}{\sin 2\beta }}$

Substitute $\sqrt{\frac{gR}{\sin 2\beta }}$ for $V_0$ in equation (II).

$t=\frac{2\left(\sqrt{\frac{gR}{\sin 2\beta }}\right)\sin \beta }{g}$

Substitute $285\text{m}$ for $R$, $9.8\text{m}/{\text{s}}^2$ for $g$ and $48.0°$ for $\beta$ in above expression.

$\begin{array}{c}t=\frac{2\left(\sqrt{\frac{\left(9.8\text{m}/{\text{s}}^2\right)\left(285\text{m}\right)}{\sin 2\left(48.0°\right)}}\right)\sin \left(48.0°\right)}{9.8\text{m}/{\text{s}}^2}\\ =8.04\text{sec}\end{array}$

Conclusion: Therefore, the time for which the ball be in flight is $8.04\sec$.

To determine

The eastward speed of the tee relative to the stars.

Answer to Problem 67CP

The eastward speed of the tee relative to the stars is $379\text{m}/\text{s}$.

Explanation of Solution

Given information: The range of the motion after hitting the ball is $285\text{m}$. The direction of initial velocity of ball is $48.0°$ above the horizontal, the radius of the circular path is $R_{\text{E}}\cos {\varphi }_{\text{i}}=\left(6.37\times {10}^6\text{m}\right)\cos 35.0°$ and the location from which the golfer tees off is $35.0°$ north latitude.

Formula to calculate the eastward speed of the tee relative to the stars is,

$V=\frac{2\pi r}{T}$

Substitute $\left(6.37\times {10}^6\text{m}\right)\cos 35.0°$ for $r$ and $24\text{h}$ for $T$ in above expression.

$\begin{array}{c}V=\frac{2\left(3.14\right)\left(\left(6.37\times {10}^6\text{m}\right)\cos 35.0°\right)}{24\text{h}\left(\frac{3600\text{sec}}{1\text{h}}\right)}\\ =379.2710\text{m}/\text{s}\\ \approx 379\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the eastward speed of the tee relative to the stars is $379\text{m}/\text{s}$.

To determine

The value by which the hole's speed exceed that of the tee.

Answer to Problem 67CP

The value by which the hole's speed exceed that of the tee is $1.19\times {10}^{-2}\text{m}/\text{s}$.

Explanation of Solution

Given information: The range of the motion after hitting the ball is $285\text{m}$. The direction of initial velocity of ball is $48.0°$ above the horizontal, the radius of the circular path is $R_{\text{E}}\cos {\varphi }_{\text{i}}=\left(6.37\times {10}^6\text{m}\right)\cos 35.0°$ and the location from which the golfer tees off is $35.0°$ north latitude.

Formula to calculate the length of the arc is,

$R=R_{\text{E}}{\varphi }^{\prime }$

Rearrange the above expression for ${\varphi }^{\prime }$.

${\varphi }^{\prime }=\frac{R}{R_{\text{E}}}\left(\frac{360°}{2\pi }\right)$

Substitute $285\text{m}$ for $R$ and $6.37\times {10}^6\text{m}$ for $R_{\text{E}}$ in above equation.

$\begin{array}{c}{\varphi }^{\prime }=\frac{285\text{m}}{6.37\times {10}^6\text{m}}\left(\frac{360°}{2\left(3.14\right)}\right)\\ =0.0025°\end{array}$

Formula to calculate the speed of the hole is,

$V_{\text{hole}}=\frac{2\pi R_E\cos \left(35.0°-{\varphi }^{\prime }\right)}{24\text{h}}$

Substitute $6.37\times {10}^6\text{m}$ for $R_{\text{E}}$, $0.0025°$ for ${\varphi }^{\prime }$ in above expression.

$\begin{array}{c}{V}_{\text{hole}}=\frac{2\left(3.14\right)\left(6.37\times {10}^6\text{m}\right)\cos \left(35.0°-0.0025°\right)}{24\text{h}\left(\left(\frac{3600\text{sec}}{1\text{h}}\right)\right)}\\ =379.2829\text{m}/\text{s}\end{array}$

Calculate the difference between the speed of tee and the speed of hole.

$\Delta V=V_{\text{hole}}-V$

Substitute $379.276\text{m}/\text{s}$ for $V_{\text{hole}}$ and $379.271\text{m}/\text{s}$ for $V$ in above expression.

$\begin{array}{c}\Delta V=379.2829\text{m}/\text{s}-379.2710\text{m}/\text{s}\\ =1.19\times {10}^{-2}\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the value by which the hole's speed exceed that of the tee is $1.19\times {10}^{-2}\text{m}/\text{s}$.

To determine

The distance to the west of the hole from the position where the ball land.

Answer to Problem 67CP

The distance to the west of the hole from the position where the ball land is $9.55\text{cm}$.

Explanation of Solution

Given information: The range of the motion after hitting the ball is $285\text{m}$. The direction of initial velocity of ball is $48.0°$ above the horizontal, the radius of the circular path is $R_{\text{E}}\cos {\varphi }_{\text{i}}=\left(6.37\times {10}^6\text{m}\right)\cos 35.0°$ and the location from which the golfer tees off is $35.0°$ north latitude.

The distance to west of the hole is given as,

$d=\Delta V\left(t\right)$

Substitute $1.19\times {10}^{-2}\text{m}/\text{s}$ for $\Delta V$ and $8.04\text{sec}$ for $t$ in above expression.

$\begin{array}{c}d=1.19\times {10}^{-2}\text{m}/\text{s}\left(8.04\sec \right)\\ =9.55\text{cm}\end{array}$

Conclusion:

Therefore, the distance to the west of the hole from the position where the ball land is $9.55\text{cm}$.