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.(adsbygoogle = window.adsbygoogle || []).push({});

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??

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Another Projectile Motion

**Physics Forums | Science Articles, Homework Help, Discussion**