At what initial speed must the basketball player in Fig. 4-50 throw the ball, at angle${\mathbf{\theta }}_{\mathbf{0}}\mathbf{=}\mathbf{55}\mathbf{°}$above the horizontal, to make the foul shot? The horizontal distances are ${\mathbf{d}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ft}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{d}}_{\mathbf{2}}\mathbf{=}\mathbf{14}\mathbf{}\mathbf{ft}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{the}\mathbf{}\mathbf{heights}\mathbf{}\mathbf{are}\mathbf{}{\mathbf{h}}_{\mathbf{1}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{ft}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{h}}_{\mathbf{2}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{ft}\mathbf{.}$

$$\begin{gathered} \mathrm{a})\mathrm{The}\mathrm{angle}\mathrm{made}\mathrm{by}\mathrm{the}\mathrm{basketball}\mathrm{above}\mathrm{the}\mathrm{horizontal},\left(\mathrm{\theta }\right)=55° \\ \mathrm{b})\mathrm{The}\mathrm{horizontal}\mathrm{distances},{\mathrm{d}}_1=1\mathrm{ft},{\mathrm{d}}_2=14\mathrm{ft} \\ \mathrm{c})\mathrm{The}\mathrm{heights}\mathrm{are},{\mathrm{h}}_1=7\mathrm{ft},{\mathrm{h}}_2=10\mathrm{ft} \end{gathered}$$

Projectile motion is the motion of a particle that is launched with an initial velocity. During its flight, the particle’s horizontal acceleration is zero and its vertical acceleration is the free-fall acceleration. (Upward is taken to be a positive direction.)

Using the equation for the projectile path in the projectile motion of the ball, we can find the initial speed of the basketball.

Formula:

The equation of the distance for the projectile path of a body,

$\mathrm{y}=\mathrm{x}{\mathrm{tan\theta }}_0-\frac{{\mathrm{gx}}^2}{2{\left({\mathrm{V}}_0{\mathrm{cos\theta }}_0\right)}^2}$ …(i)

The total horizontal displacement of basketball is given as:

$$\begin{gathered} \mathrm{x}={\mathrm{d}}_2-{\mathrm{d}}_1 \\ =13\mathrm{ft} \end{gathered}$$

The total vertical displacement of basketball is given as:

$$\begin{gathered} \mathrm{y}={\mathrm{h}}_2-{\mathrm{h}}_1 \\ =3\mathrm{ft} \end{gathered}$$

Rearranging the equation (i)for projectile path and substituting the above values, the velocity is given as:

$$\begin{gathered} {\mathrm{V}}_0=\frac{\mathrm{x}}{\mathrm{cos\theta }}\sqrt{\frac{\mathrm{g}}{2\left(\mathrm{x}\mathrm{tan\theta }-\mathrm{y}\right)}} \\ =\frac{13\mathrm{ft}}{\cos 55°}\sqrt{\frac{32.15\mathrm{ft}/{\mathrm{s}}^2}{2\left(\left(13\mathrm{ft}\right)\tan 55°-\left(3\mathrm{ft}\right)\right)}} \\ =\frac{13\mathrm{ft}}{0.574}\sqrt{\frac{32\mathrm{ft}/{\mathrm{s}}^2}{31.13\mathrm{ft}}} \\ =22.98\mathrm{ft}/\mathrm{s} \\ \approx 23\mathrm{ft}/\mathrm{s} \end{gathered}$$

Hence, the value of the initial speed of the ball is 23ft/s.