Projectile Que HELP- Finding Launch Angles

[lavenderbaby]

Homework Statement

A fire fighter directs a stream of water from a fire hose into the window of a
burning building. The window is 20.0 m above the level of the nozzle and 60.0 m
away. The hose produces a jet of water at a constant nozzle speed of V0 = 30.0 m/s.
The fire fighter can control the angle θ that the water is aimed at by moving the
nozzle with her hands.

(a) The fire fighter discovers that there are exactly two angles θ that she can hold
the hose for which the water gets into the window. What are the two angles?

Homework Equations

Y1=Y0 + V0y(t) - g/2(t)^2
X1=X0 + V0x(t)
V0x = Vcos(θ)
V0y = Vsin(θ)
etc.

The Attempt at a Solution

Before getting into any calculations, I know that the angles are θ and (90-θ). That is, at these two angles, the firefighter will shoot the same height. Once I find out one angle, I will be able to get the other. But the problem is that I can't find even one θ.

First, I wrote down equations of projectile motion in the x-direction and in the y-direction. But I ended up getting a complicated equation involving both cos^2(θ) and tan(θ) that I can't solve. Then, I attempted the question in a different way. I tried to find the time it takes the water stream to reach the maximum height, and the time it takes the water stream to travel from the maximum height to the window. As you can see on the sheet, my equations involve both θ and t.

Is there a way that I can get rid of the t variable? Or what else should I do?

Please shed light! I've spent two hours writing over seven sheets w/ no result.

Thanks in advance!

 

[tiny-tim]

Welcome to PF!

Hi lavenderbaby! Welcome to PF!

(try using the X² tag just above the Reply box )

… But I ended up getting a complicated equation involving both cos^2(θ) and tan(θ) that I can't solve. Then, I attempted the question in a different way. I tried to find the time it takes the water stream to reach the maximum height, and the time it takes the water stream to travel from the maximum height to the window. …

Yes, you got A/cos²θ + Btanθ + C = 0 …

just multiply it throughout by cos²θ,

then use standard trigonometric identities to write it as a combination of cos2θ and sin2θ,

then use another trigonometric identity to write that as cos(2θ + λ), for some λ.

(and no, finding the maximum height is never going to help)

 

[lavenderbaby]

Hi tiny-tim, thanks for your prompt reply! But in order to write it as a combo of cos and sin, do i need to rewrite the tan in terms of sin/cos first? I'm not really sure how that is going to give me a combo of sin and cos...any more hints?

Hi lavenderbaby! Welcome to PF!

(try using the X² tag just above the Reply box )


Yes, you got A/cos²θ + Btanθ + C = 0 …

just multiply it throughout by cos²θ,

then use standard trigonometric identities to write it as a combination of cos2θ and sin2θ,

then use another trigonometric identity to write that as cos(2θ + λ), for some λ.

(and no, finding the maximum height is never going to help)

 

[tiny-tim]

Not cosθ and sinθ, cos2θ and sin2θ.

cos²θtanθ = … ?

 

[BoggyP]

hi! I too require help.

I got as far as: 5+5cos(2$$\Theta$$) = 15sin(2$$\Theta$$) + a

I'm not sure how to solve this :/