Ah okay I didn't know that
So if I leave the equation in the form:
v^2 = vo^2 + 2ax
I get:
(-55sinθ - 32.174*(35/55cosθ))^2 = (55sinθ)^2 + 2*(-32.174)(-20)
Which turns into:
419.2cos(θ)^2+2252.18cosθsinθ-1286.96 = 0
Also when I try using the equation x = x0 + v0t + 1/2at^2 I get a...
Thanks :)
I see what I did wrong there, the vertical velocity is actually:
v = -55sinθ1 + (-32.174)(35) / 55cosθ1
Therefore:
v= v0 + sqrt(2ax)
(-55sinθ1 + (-32.174)(35/55cosθ1)) = -55sinθ1 + sqrt(2(-32.174)(-20))
I'm still doing this for below the horizon though so when I do algebra for the...
Homework Statement
The fireman wishes to direct the flow of water from his hose to the fire at B.
Determine two possible angles θ1 and θ2 at which this can be done. Water flows from the hose at v = 55 ft/s.
There is no air friction.
Homework Equations
v = v0 + at
x = x0 + vt
v2 = v^{2}_{0}...