Finding Values for Thrown Stone from 45m Building

[aquapod17]

Homework Statement

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 ______°

Homework Equations

v_ix = v_icostheta
v_iy = v_isintheta

d_x = v_xt
d = v_it + 1/2at²

The Attempt at a Solution

I tried to find v_ix and v_iy in terms of vi and then used d_x = v_xt to get t as 30.0m/.899v_i but I don't know if (1) I'm doing it right and (2) where to go from here.

 

[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.