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$\begin{array}{l}{12-93 . \text { A golf ball is struck with a velocity of } 80 \mathrm{ft} / \mathrm{s} \text { as }} \\ {\text { shown. Determine the distance } d \text { to where it will land. }}\end{array}$

$v=80 f t / s$

$(\stackrel{+}{\rightarrow}){S}=S_0+V_0 t$

$x={x}_0+{v_{0}}_{x} t$

$x=d \cos 10^{\circ}=0+80 \cos 55^{\circ} t$

$d \cos 10^{\circ}=80 \cos 55^{\circ} t \Rightarrow (1)$

$(+ \uparrow) \ y=y_0+(v_0)_y t - \frac{1}{2} g t^2$

$d \sin 10=0+80 \mathrm{sin} 55^{\circ} t-\frac{1}{2}gt^2$

$d \sin 10=80 \sin 55^{\circ}+\frac{1}{2}(32.2)t^2$

$\begin{array}{l}{12-98 . \text { Determine the horizontal velocity } v_{A} \text { of a tennis }} \\ {\text { ball at } A \text { so that it just clears the net at } B \text { . Also, find the }} \\ {\text { distance } s \text { where the ball strikes the ground. }}\end{array}$

$\left.\begin{array}{l}{v_{A}=?} \\ {s=?}\end{array}\right\} V. Motion$

$\left(v_{0}\right) y=0 \ { \overrightarrow {A \rightarrow B}} (s_0)_y=7.5 ft \ , s_y=3 ft$

$(+\uparrow)\Rightarrow ∵ s_ y=(s_0)_ y+(v_{0})_ y t+\frac{1}{2}(a_c)_ y t^{2}$

$3=7.5+0+ \frac{1}{2}(-32.2)t_1^{2}$

$t_{1}=0.5287 s\Rightarrow (1)$

$\left(S_{0}\right)_ y=7.5 f t \ ,S_y=0$

${0=7.5+0+\frac{1}{2}(-32.2) t_{2}^{2}}$

$t_2=0.6825s$

H. Motion

$\left(v_{0}\right)_{x}=v_{A} \overrightarrow {A \longrightarrow B}\left(S_{0}\right)_{x}=0, \ S_{x}=21 ft$

$t-t_{1}=0.5287s$

$S_{x}=\left(s_{0}\right)_ x+\left(v_{0}\right)_{x} t$

$21=0+v_A (0.5287)$

$v_{A}=39.72 \mathrm{ft} / \mathrm{s}≃39.7 \mathrm{ft} / \mathrm{s}$

$A \rightarrow C \Rightarrow (s_0)_x=0, \ s_x=(21+s) ft$

$t=t_{2}=0.6825S$

$S_{x}=\left(S_{0}\right)+\left(V_{0}\right)_{x} t$

$21+S=0+39.72(0.6825)$

$S=6.11 \mathrm{Ft}$