# Another Projectile Motion

1. Oct 15, 2005

### NoMeGusta

Basketball player is 2m tall. He shoots at 40 degrees over a distance of 10m into a basketball net that is 3.05m. Whats the initial velocity.

d=10m
Theta=40 degrees

Vxi = Vi cos(40) Ax=0 xi=0 xf=10
Vyi= Vi sin(40) Ay=-9.80 yf=1.05 (difference between player and net)

1. First I wanted to solve for time

Xf = Xi + Vxi*t + (1/2)Ax*(t^2)
10 = 0 + (Vi*cos(40))t + (1/2)(0) (t^2)
10 = (Vi*cos(40))t

t = (Vi*cos(40))/ 10

2. Then, I'd use the Yf formula to plug my new time into to get Vi since Vyi= Vi sin(40)

Yf = Yi + Vyi*t + (1/2)Ay*(t^2)
1.05=0 + (Vi*sin(40))t - (1/2)(-9.80)(t^2)

---then substituting t into the 1.05=0 + (Vi*sin(40))t + (1/2)(-9.80)(t^2)
and after all the algebraic dust settled I got ---

Vi = 7.159.... the problem is the book says 10.7 m/s. Where did I go wrong??

Last edited: Oct 15, 2005
2. Oct 15, 2005

### ehild

You have used too many "-"-s.
ehild

3. Oct 15, 2005

### Fermat

For the time, it should be,

t = 10/(Vi*cos(40))

4. Oct 15, 2005

### NoMeGusta

that was a typo

5. Oct 15, 2005

### NoMeGusta

Holy crap... i'm staring at my page of work and you are sooooo right. Let me try that and see if it works now.

6. Oct 15, 2005

### NoMeGusta

It worked... It worked!!!

That was the fix. I had the equation for t wrong. I got the 10.7 book answer now. Thank you sooooooooooo much!!!