Answer to Question #96694 in Mechanics | Relativity for Abru

Question

A basketballer is trying to make a shot in an international game from the 3 point line (6.75 m
away from the hoop). The basketball hoop is 3.05 m above the floor, the basketballer’s hands
are 1.95 m above the floor when they release the ball at an angle of 39° above the horizontal.
What velocity would the ball have to have been thrown with to reach and fall down into the
hoop? And if the ball has a mass of 625 g what was the ball’s maximum potential energy during
the throw?

Answer 1

The equation of trajectory of the ball when it moves as projectile

$$y=y_0+\tan(\theta)\cdot x-\frac{g}{2v_0^2\cos^2(\theta)}\cdot x^2$$

In our case we have an equation

$$3.05=1.95+\tan(39^{\circ})\cdot 6.75-\frac{9.8}{2v_0^2\cos^2(39^{\circ})}\cdot 6.75^2$$

So, the initial velocity

$$v_0=9.2 \:\rm m/s$$

The maximum height

$$y_{max}=y_0+\frac{v_0^2\sin^2(\theta)}{2g}$$

$$=1.95+\frac{9.2^2\sin^2(39^{\circ})}{2\times 9.8}=3.66\:\rm m$$

So, the ball’s maximum potential energy during the throw

$$E_{max}=mgy_{max}$$

$$=0.625\:\rm kg\times 9.8\: N/kg\times 3.66\: m=22.4\:\rm J$$