Question

The free throw line in basketball is $\mathbf{4}\mathbf{.}\mathbf{57}\mathit{m}\mathbf{(}\mathbf{15}\mathbf{}\mathit{f}\mathit{t}\mathbf{)}$from the basket, which is$\mathbf{3}\mathbf{.}\mathbf{05}\mathit{m}\mathbf{\left(}\mathbf{10}\mathit{f}\mathit{t}\mathbf{\right)}$ above the floor. A player standing on the free throw line throws the ball with an initial speed of$\mathbf{7}\mathbf{.}\mathbf{15}\mathit{m}\mathbf{/}\mathit{s}$ releasing it at a height of$\mathbf{2}\mathbf{.}\mathbf{44}\mathit{m}\mathbf{(}\mathbf{8}\mathbf{}\mathit{f}\mathit{t}\mathbf{)}$ above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error. Explicitly show how you follow the steps involved in solving projectile motion problems.

Short answer

The ball must be thrown at an angle of degrees above the horizontal to exactly hit the basket.

Step-by-step solution

Stating given data

The change in displacement in y direction is meter.

The acceleration in the above case is -9.8 m/s².

The initial velocity in the x direction is

$$\begin{gathered} \cos \theta =\frac{a}{h} \\ \cos \theta =\frac{V_{ix}}{V_i} \\ {V}_{ix}=8.15\cos \theta \end{gathered}$$

The initial velocity in the y direction will be

Calculating angle at which ball should be thrown

The velocity in the x frame will be constant at all times.Considering the initial velocity

$$\begin{gathered} \overline{V}=\frac{x}{t} \\ 8.15\cos \theta =\frac{4.57}{t} \\ t=\frac{4.57}{8.15\cos \theta } \end{gathered}$$

The time taken is used from the above equation.

Putting the values in the equation of motion, we have

$$\begin{gathered} ∆Y=V_{iy}t+\frac{1}{2}{a}_y{t}^2 \\ 0.61=(8.15\sin \theta )t+\frac{1}{2}(-9.8)t^2 \end{gathered}$$

Substituting the value of t

$$\begin{gathered} .61=\left(V_i\sin \theta \right)\left(\frac{4.57}{V_i\cos \theta }\right)+\frac{1}{2}(-9.8){\left(\frac{4.57}{8.15\cos \theta }\right)}^2 \\ 0.61=4.57\tan \theta -4.9{\left(\frac{4.57}{8.15\cos \theta }\right)}^2 \\ 0.61=4.57\tan \theta -1.54({\tan }^2\theta +1) \\ 1.54{\tan }^2\theta -4.57\tan \theta +2.15=0 \end{gathered}$$

(We can use a calculator to solve the above quadratic equation.)

Hence the value of the tangent is

The player uses the larger value of the angle; so, we can choose the angle of 67.1 degrees.