Homework Help: 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!!!