Projectile Motion (Angles of projection problem)

[CaptainSMASH]

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} + 2ax

The Attempt at a Solution

Going to try and find the first angle (below the x-axis).

Components of the velocity:
V_x = 55cosθ₁
V_y = 55sinθ₁

Horizontal velocity is constant therefore we can find an equation for time:
x = x₀ + vt
35 = 0 + (55cosθ₁)t
t = 35/55cosθ₁

Sub this into an equation for vertical velocity to find v final:
v = v₀ + at
v = 55sinθ₁ + (-32.174)(35/55cosθ₁)

Sub this velocity into another equation for vertical velocity:
v² = v^{2}_{0} + 2ax
v= v₀ + \sqrt{2ax}
(55sinθ₁ + (-32.174)(35/55cosθ₁)) = 55sinθ₁ + \sqrt{2(-32.174)(-20)}
...use algebra to find the angle

I'm not sure if this gives the right answer for the first angle, I hope it does, and I'm unsure as to how to get the second angle. Do I break the question for θ₂ into two parts? One equation for the vertical rise and then a second for the fall? I've been figuring this question out more and more as I typed this out (I hope) but it's been bugging me since yesterday, I'm hoping to get some assurance / help from you guys.

Thanks in advance.

 

[rl.bhat]

Hi CaptainSMASH, welcome to PF.

v = 55sinθ1 + (-32.174)(35) / 55cosθ1)

Check this step. vo*sinθ and g are in the same direction in the first case. You have complicated calculation for the hose directed above the horizon.

 

[CaptainSMASH]

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 angle the v's should still cancel.

When I do algebra I get:

θ₁ = cos-1(sqrt(2(-32.174)(-20))/(35/55)

..I am bad at this forum code, I simplified it.

Are there any other issues with my method to get θ₁?

 

[rl.bhat]

One more thing.
v^2 = vo^2 + 2ax.
You can't take the square root of above equation and write
v = vo + sqrt(2ax)

 

[CaptainSMASH]

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 similar ugly polynomial equation. I feel like I shouldn't need a graphing calculator to do this question..

 

[rl.bhat]

To avoid all these calculations, use the following formula.
-y = -vo*sinθ*t - 1/2*a*t^2
-y = -vosinθ*x/vo*cosθ - 1/2*α*x^2/(vo*cosθ)^2
-y = -x*tanθ - 1/2*a*x^2*sec^2(θ)/vο^2
y = x*tanθ + 1/2*a*x^2*(1 + tan^2(θ))/vο^2
Substitute the values and solve the quadratic to get two values of angle. One above the horizon and the other below the horizon.