# Projectile Motion!

1. Nov 17, 2008

### aquapod17

1. The problem statement, all variables and given/known data
Suppose a stone is thrown from a 45.0 m tall building at an angle 26.0° below the horizontal. If it strikes the ground 30.3 m away, find the following values. (a) time of flight
__________s

(b) initial speed
_________ m/s

(c) speed and angle of the velocity vector with respect to the horizontal at impact
_________m/s ______°

2. Relevant equations
vix = vicostheta
viy = visintheta

dx = vxt
d = vit + 1/2at2

3. The attempt at a solution
I tried to find vix and viy in terms of vi and then used dx = vxt to get t as 30.0m/.899vi but I don't know if (1) I'm doing it right and (2) where to go from here.

2. Nov 17, 2008

### LowlyPion

You should get 2 equations:

30.3 = V*Cosθ * t

45 = V*Sinθ * t + 1/2 * g* t2

V * t = 30.3/Cosθ

==> 45 = Sinθ * (30.3/Cosθ ) + 1/2*g*t2

1/2*g*t2 + 30.3*Tanθ - 45 = 0

Solve the quadratic for t and then figure V from either equation.

For final velocity Vfy = V*Sinθ + g*t Your Vfx is still V*Cosθ so you should be done.