Question

question 2 of 15
in the sport of billiards, event organizers often remove one of the rails on a pool table to allow players to measure the speed of their break shots (the opening shot of a game in which the player strikes a ball with his pool cue).

the top of a pool table is a height ( h = 2.85 ) ft from the floor.

if a player’s ball lands a distance ( d = 14.75 ) ft from the table edge, what is her break shot speed ( v_0 )? (quad v_0 = ) (\boxed{quad}) mph
at what speed ( v ) did her pool ball hit the ground? (quad v = ) (\boxed{quad}) mph

Explanation:

Step1: Analyze vertical motion (free fall)

The vertical motion of the ball is a free - fall motion with initial vertical velocity \(v_{0y} = 0\space ft/s\), acceleration \(a = g=32.2\space ft/s^{2}\) (acceleration due to gravity) and displacement \(y=-h=- 2.85\space ft\) (taking downwards as negative). We use the equation \(y = v_{0y}t+\frac{1}{2}at^{2}\).
Since \(v_{0y} = 0\), the equation simplifies to \(y=\frac{1}{2}at^{2}\), so \(t=\sqrt{\frac{2|y|}{a}}\).
Substituting \(|y| = h = 2.85\space ft\) and \(a = 32.2\space ft/s^{2}\), we get \(t=\sqrt{\frac{2\times2.85}{32.2}}=\sqrt{\frac{5.7}{32.2}}\approx\sqrt{0.177}\approx0.421\space s\).

Step2: Analyze horizontal motion (uniform motion)

The horizontal motion of the ball is a uniform motion with constant velocity \(v_{0x}=v_{0}\) (since there is no horizontal acceleration). The horizontal displacement \(x = d = 14.75\space ft\) and \(x=v_{0x}t\), so \(v_{0}=\frac{d}{t}\).
Substituting \(d = 14.75\space ft\) and \(t = 0.421\space s\), we get \(v_{0}=\frac{14.75}{0.421}\approx35.03\space ft/s\).
To convert from \(ft/s\) to \(mph\), we use the conversion factor: \(1\space ft/s=\frac{3600}{5280}\space mph\approx0.6818\space mph\).
So \(v_{0}=35.03\times0.6818\approx23.9\space mph\).

Step3: Find the vertical component of velocity when hitting the ground

We use the equation \(v_{y}=v_{0y}+at\). Since \(v_{0y} = 0\), \(v_{y}=at\).
Substituting \(a = 32.2\space ft/s^{2}\) and \(t = 0.421\space s\), we get \(v_{y}=32.2\times0.421\approx13.56\space ft/s\).

Step4: Find the speed when hitting the ground

The speed \(v\) when hitting the ground is the magnitude of the velocity vector, given by \(v=\sqrt{v_{0x}^{2}+v_{y}^{2}}\). Since \(v_{0x}=v_{0}\approx35.03\space ft/s\) and \(v_{y}\approx13.56\space ft/s\),
\(v=\sqrt{(35.03)^{2}+(13.56)^{2}}=\sqrt{1227.1 + 183.9}\approx\sqrt{1411}\approx37.56\space ft/s\).
Converting to \(mph\): \(v = 37.56\times0.6818\approx25.6\space mph\).

Answer:

Break shot speed \(v_{0}\approx\boldsymbol{23.9}\space mph\) (or more precisely, after re - calculating the conversion:
First, \(v_{0}=\frac{d}{\sqrt{\frac{2h}{g}}}\)
\(v_{0}=\frac{14.75}{\sqrt{\frac{2\times2.85}{32.2}}}=\frac{14.75}{\sqrt{\frac{5.7}{32.2}}}=\frac{14.75}{\sqrt{0.177}}\approx\frac{14.75}{0.4207}\approx35.06\space ft/s\)
\(35.06\space ft/s\times\frac{3600\space s}{1\space h}\times\frac{1\space mile}{5280\space ft}=35.06\times\frac{3600}{5280}\approx35.06\times0.6818\approx23.9\space mph\)

Speed when hitting the ground:
\(v_{y}=gt = 32.2\times\sqrt{\frac{2\times2.85}{32.2}}=\sqrt{2\times32.2\times2.85}=\sqrt{183.42}\approx13.54\space ft/s\)
\(v=\sqrt{v_{0}^{2}+v_{y}^{2}}=\sqrt{(35.06)^{2}+(13.54)^{2}}=\sqrt{1229.2 + 183.3}=\sqrt{1412.5}\approx37.58\space ft/s\)
\(37.58\times\frac{3600}{5280}\approx37.58\times0.6818\approx25.6\space mph\)

So \(v_{0}\approx\boldsymbol{24}\space mph\) (rounded) and \(v\approx\boldsymbol{26}\space mph\) (rounded) or more accurately \(v_{0}\approx23.9\space mph\) and \(v\approx25.6\space mph\))