what is the release speed of the ball 4 player to make shot?

Question

player= 2.05m tall, distance to basket= 6.02m, height of basket from floor= 3.05m, angle ball launched at= 25 degrees level with players head. if you can show me how you come up w/answer. thanks

Answer 1

Wow... another nice question...

Let's start with the range (Horiz distance) of 6.02 m
vX = 6.02 / time

We know that vX = v * cos θ, so
v * cos 25 = 6.02 / t
v = 6.02 / (t * cos 25)
v = 6.64 / t OR
t = 6.64 / v

Now, let's think vertical.
vY = v * sin 25

Time to reach max height
t = (vF - vI)/g
= (0 - (v * sin 25) / -9.81
= v * sin 25 / 9.81
= v * 0.0431

Note that this time is not the same as the time used for the range.
We need to find the time from the max height to the basket (on way down)

Let's find max height
d = (vF^2 - vI^2) / (2 * -g)
= (0 - (v * sin 25)^2) / (2 * -9.81)
= 0.179 * v^2 / 4.905
= 0.00910 * v^2

Now, calculate the time to fall from max height to basket.
This is a distance of max height minus the 1 meter difference b/w player and net
t = sqrt(2 * d / g)
= sqrt(2 * (0.00910 * v^2 - 1) / 9.81)

so total time (vertically) is = 0.0431 * v + sqrt(2 * (0.00910 * v^2 - 1) / 9.81)

Now, with a bit of iteration skills, we get finally
v = 10.94 m/s

WHEW!