The 0.1-lb golf ball is struck by the club and then travels along the trajectory shown. De...

Accordingly, we have from (1)

\(500=0+0.866 v \times t\)

\(t=\frac{500}{0.866 v}\)

\(t=\frac{577.367}{v} \quad \ldots \ldots . .\)

We have vertical motion of the golf ball as

\(s_{y}=\left(s_{0}\right)_{y}+\left(v_{0}\right)_{y} t+\frac{1}{2} a_{y} t^{2}\)

Where, we have \(s_{y}=0 ;\left(s_{0}\right)_{y}=0 ; a_{y}=-g\)

By substituting the values of the parameters into (3), we have

\(0=0+v \sin 30^{\circ} \times\left(\frac{577.367}{v}\right)+\frac{1}{2} \times(-32.2) \times\left(\frac{577.367}{v}\right)^{2}\)

\(0=288.6835-\frac{5366977.71}{v^{2}}\)

\(v^{2}=\frac{5366977.71}{288.6835}\)

\(v=136.35 \mathrm{ft} / \mathrm{s}\)

From the principle of linear impulse and momentum, we have

Initial speed of the golf ball is

Speed of the golf ball, is

By substituting the values of the parameters into \((4)\), we have

\(\left(\frac{0.1}{32.2}\right) \times 0+F_{\text {avg }} \times(0.0005)=\left(\frac{0.1}{32.2}\right) \times 136.35\)

\(0.0005 F_{\text {ang }}=0.42345 \mathrm{lb}\)

\(F_{a y g}=846.9 \mathrm{lb}\)

Average impulsive force the club imparts on the ball is \(\quad F_{\text {avg }}=846.9 \mathrm{lb}\)