Projectile Motion Angles Question

[iurod]

Homework Statement

A fire hose held near the ground shoots water at a speed of 6.8m/s. At what angle(s) should the nozzle point in order that the water land 2.0m away? Why are there two different angles?

Homework Equations

v = v(initial) +at
x = x(initial) + v(initial)t + .5at^2
v^2 = v^2(initial) + 2a(x - x(initial)
x=vt

The Attempt at a Solution

I first drew a right triangle and by the problem i was only given the hypotenuse (6.8). Knowing that the adjacent line is the initial velocity in the positive x direction and that the opposite line is the initial velocity in the y positive direction, i tried to solve for theata that way.

So I did:

Initial Velocity in the positive X direction is:

X= X(initial) + V(initial)(t) + .5(a)(t^2)
2= 0 + (6.8)(cos theta)(t) + 0
2= (6.8)(cos theata)(t)


Initial velocity in the positive Y direction is:

Y= Y(initial) + V(initial)(t) + .5(a)(t^2)
0= 0 + (6.8)(sin theta)(t) + (-4.9)t^2
0= (6.8)(sin theta)(t) + (-4.9)t^2

Once I got to this point I had no idea what to do... any guidance from here would be greatly appreciated. I've never taken Physics before and took geometry around 10 years ago so this has all been erased from my memory..

Thanks

 

[graphene]

from your first equation, get t in terms of cos theta.
substitute for t in the second equation.
you should get a quadratic in tan theta.

 

[iurod]

Ok, so if I'm following correctly; solving the first equation for t is:

t=(2/6.8cos theta)

Substitute t in equation 1 for that in equation so:

0 = (6.8sin theta)(2/6.8cos theta) + -4.9(2/6.8 cos theta)²

How do I make a quadratic and solve for something like this?

 

[inky]

Ok, so if I'm following correctly; solving the first equation for t is:

t=(2/6.8cos theta)

Substitute t in equation 1 for that in equation so:

0 = (6.8sin theta)(2/6.8cos theta) + -4.9(2/6.8 cos theta)²

How do I make a quadratic and solve for something like this?

Here have a look very simple method.
t=2/(6.8cos θ) ...(1)
For y-axis
0=(6.8sinθ)t - 4.9t^2
t=(6.8sin θ)/4.9...(2)
Eqn.(1)=Eqn.(2)
2/(6.8cos θ) =( 6.8sin θ)/4.9
2sin θ cos θ=4(4.9)/(6.8)^2
sin2 θ=4(4.9)/(6.8)^2
θ=...

 

[iurod]

Here have a look very simple method.
t=2/(6.8cos θ) ...(1)
For y-axis
0=(6.8sinθ)t - 4.9t^2
t=(6.8sin θ)/4.9...(2)
Eqn.(1)=Eqn.(2)
2/(6.8cos θ) =( 6.8sin θ)/4.9
2sin θ cos θ=4(4.9)/(6.8)^2
sin2 θ=4(4.9)/(6.8)^2
θ=...

Ok I followed you up to here:
t=2/(6.8cos θ) ...(1)
For y-axis
0=(6.8sinθ)t - 4.9t^2
t=(6.8sin θ)/4.9...(2)
Eqn.(1)=Eqn.(2)


If I set Eqn. 1 = Eqn. 2

(2/6.8 cosθ) = (6.8 sinθ)t + -4.9t²

now how do I get rid of the sin and cos, sorry I'm new to this and don't fully understand how to do this.

 

[inky]

Ok I followed you up to here:
t=2/(6.8cos θ) ...(1)
For y-axis
0=(6.8sinθ)t - 4.9t^2
t=(6.8sin θ)/4.9...(2)
Eqn.(1)=Eqn.(2)


If I set Eqn. 1 = Eqn. 2

(2/6.8 cosθ) = (6.8 sinθ)t + -4.9t²

now how do I get rid of the sin and cos, sorry I'm new to this and don't fully understand how to do this.

You have to see my method. Use 2sinθcosθ =sin 2θ formula.

 

[rl.bhat]

0 = (6.8sin theta)(2/6.8cos theta) + -4.9(2/6.8 cos theta)2

Rewrite this equation as

0 = 2tan(θ) - [4.9*4/(6.8)^2]sec^2(θ)

0 = 2tan(θ) - [4.9*4/(6.8)^2][1 + tan^2(θ)]

0 = 2tan(θ) - [4.9*4/(6.8)^2] - [4.9*4/(6.8)^2][tan^2(θ)

Now solve the quadratic to find two angles.

 

[inky]

0 = (6.8sin theta)(2/6.8cos theta) + -4.9(2/6.8 cos theta)2

Rewrite this equation as

0 = 2tan(θ) - [4.9*4/(6.8)^2]sec^2(θ)

0 = 2tan(θ) - [4.9*4/(6.8)^2][1 + tan^2(θ)]

0 = 2tan(θ) - [4.9*4/(6.8)^2] - [4.9*4/(6.8)^2][tan^2(θ)

Now solve the quadratic to find two angles.

I explain simple method.

t=2/(6.8cos θ) ...(1)

t=0 at initial point, t= 2/(6.8cos θ) at final point

For y-axis
0=(6.8sinθ)t - 4.9t^2
t=(6.8sin θ)/4.9...(2)
Eqn.(1)=Eqn.(2)
2/(6.8cos θ) =( 6.8sin θ)/4.9
2sin θ cos θ=4(4.9)/(6.8)^2
sin2 θ=4(4.9)/(6.8)^2=0.4239
2θ= 25.81 deg. (or) 180-25.81=154.19 deg.
θ = 12.54 deg. (or) 77.46 deg.

 

[iurod]

Thank you so much Inky, I now get it... I forgot some of the functions and seeing it step by step brought them back to me.

Thanks again!