Question 4.3: A basketball player throws a ball at an angle θ0 = 60° above......
A basketball player throws a ball at an angle θ_{0} = 60° above the horizontal, as shown in the Fig. 4.11. At what speed must he throw the ball to score?
In terms of the unknown speed v_{0} , we must first use the horizontal Eq. 4.18
x=v_{x\circ }t ⇒ x=\left(v_{\mathrm{o}}\cos\theta_{\mathrm{o}}\right)t (4.18)
to find the time needed for the ball to reach the net after moving a horizontal distance of 4 m. Thus, after omitting the units temporarily since they are consistent, we get:
x=(v_{o}\cos\theta_{o})t\;\;\Rightarrow\;\;\;4=v_{o}(\cos 60°)t=0.5v_{o}t\;\;\Rightarrow\;\;\;t=\frac{8}{v_{o}}Since y is up, then a = −g = −9.8 m/s² during ascending and descending motions. Also, choosing the origin of the xy plane to be at the player’s hands makes y_{0} = 0. When the ball reaches the net, it gains vertical height y = 1 m.
Then, by using Eq. 4.20 we get:
y=v_{y o}t-{\frac{1}{2}}g t^{2} ⇒ y=(v_{\mathrm{o}}\sin\theta_{\mathrm{o}})t-{\textstyle\frac{1}{2}}g t^{2} (4.20)
y=(v_{o}\sin\theta_{o})t-{\frac{1}{2}}g t^{2}~~\Rightarrow~~1=(v_{o}\sin60^{\circ}){\frac{8}{v_{o}}}-{\frac{1}{2}}\times9.8\times\left({\frac{8}{v_{o}}}\right)^{2}or,
1=6.93-{\frac{313.6}{v_{o}^{2}}}Thus, solving for v_{o} and taking the positive root gives:
v_{\mathrm{o}}={\sqrt{\frac{313.6}{6.93-1}}}=7.27{\mathrm{ \ m/s}}