Dynamics: solve both trajectories

[600burger]

Hey all,

Having trouble solving this one. Class is dynamics. 2D, constant accel.

Shooting a cannon at a point above the initial. Inital velocity is 400m/s

Cannon is at point A (origin) and we're shooting at point B @ (5000m, 1500m)

I'm asked to find the two thetas that satisfy.

So i got this so far, basicly the givens.

In the x:

ax = 0
vx = v0*cos(Θ)
x = v0*cos(Θ)*t

In the y:

ay = -g
vy = v0*sin(Θ)-gt
y = v0*sin(Θ)*t-(1/2)gt2

I tried to eliminate t is the y position equation by

1) solving x for t
2) subing into y(Θ,t) for t to get y(Θ)
3) solving for Θ

I can't get through solving for Θ. Can I use the quadradic equation on y(Θ) to solve for Θ? How do i do that?

 

[radou]

... Can I use the quadradic equation on y(Θ) to solve for Θ? How do i do that?

Yes, you should be able to solve the quadratic equation for $$\Theta$$, after plugging in the values of displacement 5000 m and 1500 m.

 

[600burger]

Thats what my prof was saying, but I am not sure how that's done? cause it some up to like


-a*sec(Θ)^2 +b*tan(Θ) - c = 0

so does it matter that Θ is in sec^2 for the a and tan for b?

 

[600burger]

**In the voice of reason**

Learn your trig identities backwards and forwards!

*echo 7th grad algebra teachers voice

**End voice**


sec(Θ)^2 = 1 + tan(Θ)^2

Sub and solve. Still an ugly problem, but I'm gald I figured it out.