- #1

- 224

- 0

The device will consist of a spring inside a pipe. At the top of the spring (where the projectile will be) there will be a small flat piece of wood with a rope going through it and through the spring to the other end. The rope will come out of the other end and go under a pully supported on both sides and I will pull the rope to contract the spring and then let it go.

Here are my calculations:

the basketball hoop is about 10 ft high so that's about 3 m. I did

Vf^2-Vi^2=2ad where Vf = 0, Vi = initial vertical speed, a = 9.8 m/s^2, and d = 3m

I came up with Vi (initial vertical speed) =

**7.668 m/s**

I then used Vf-Vi = at to find the time in flight where Vf = 0, Vi = initial vertical speed, a = 9.8 m/s, and t = half of the flight time.

Total time ended up being about

**1.56s**

The device will be placed behind the foul line which is about 15 ft away from the basket or about 4.57m which means that the range of the device should be 2 * 4.57m = 9.14 m.

I then used V = d/t to find the horizontal initial velocity and came up with:

V = 9.14m/1.56s =

**5.859 m/s**

From the initial horizontal velocity and initial vertical velocity, I used vectors to find the initial velocity and the angle and came up with:

Vi = sqrt(Vih^2+Viv^2) =

**9.65 m/s**

angle = tan^(-1)(7.668/5.859) =

**52.6 degrees**.

So far, I found the angle at which the pipe should be placed and the initial velocity of the projectile, but to buy the string I have to find the string constant and here's what I did to find it:

F = kx.

KE (kinetic energy) = int(kx dx) = (kx^2)/2

KE = (1/2)(mv^2)

(kx^2)/2 = (mv^2)/2

**k = (m*v^2)/(x^2)**where m is the mass of the projectile, v is the initial velocity I came up with (9.65 m/s), and x is the displacement of the spring. I can control the displacement but it will probably be about .5 m so that gives me a spring constant of about 14.8996 Kg/s^2 (according to my last equation). The units are wrong so I probably did something wrong, anyone know where I went wrong?