Question

A coach is desperate to find a sol'n to the poor performance of his b-ball team at the free-throw line. He wants a fool-proof equation that defines all successful free-throws in terms of the initial velocity, Vi, and theta. The rim of the basket is 3.05 m high and the foul line is 4.19 m from the centerline of the rim. Assume the ball is launched from a height of 1.83 m. (g=-9.80 m/s^2)

Determine an equation for making a free-throw.

Answer 1

Ah, foul line is the same as the free-throw line.

Answer 2

$$
\large
h={v_i^2\sin^2\theta_i\over 2g}
$$
\(\large v_i\) is initial velocity, \(\large \theta_i\), is initial angle.

Note there are an infinite number of initial velocities and angles that will get the ball to the top of the rim.

Answer 3

But, since we are also given the horizontal distance \(R\) to the basket, we can use it to solve for a unique velocity and angle:
$$
\large
R={v_i^2\sin2\theta_i\over g}
$$
Two equations, two unknowns. Solvable.

Answer 4

Haha, but we have to derive those equations from the basic 5 kinematics equations.

Answer 5

And that's where I'm lost. I end up with something under a root in the equation and I'm not sure where to go from here.

Answer 6

you can use two equations.
1. equation of motion in x direction
\[x=v_{0x}t=tv_0\cos{\theta}\]
2. equation of motion in y
\[y=y_0+v_{0y}t-gt^2/2\]
take t from 1st insert into 2nd:
\[y=y_0+v_0\sin{\theta}(x/v_0\cos{\theta})-g(x/v_0\cos{\theta})^2/2\]
here y=3.05; x=4.19;y_0=1.83
you will get equation which will include relation between v_0s and thetas