Solving for Launch Angle Using Vo

[Alex T Weliever]

Homework Statement

Find the angle at which the projectile is launched.
(Hint: use tan^2+1=sec^2)

Initial Net Velocity=430 m/s
The projectile lands perfectly on the edge of a cliff 80 meters high and 175 meters away.

Homework Equations

tan^2+1=sec^2
Kinematics, i.e.-
v=vo+gt
dy=vo(dt)+.5(-10)dt^2
v^2=Vo^2+2(-10)dy
and so on.

The Attempt at a Solution

I've gotten as far as listing my knowns and solving out for a few things implicitly, nothing concrete really solved.

Vox=430cos theta
Voy=430 sin theta
I believe the following is true?
43*sin theta=t(peak)
86*sin theta=t(final, not taking the cliff ledge into account)

What I'm struggling with is just beginning, really. I don't want someone to tell me the answer, just guide me to what I need to start with. I started by focusing on the Vox/Voy triangle to solve for velocities, but then I realized with that information I could solve for the angle automatically, and that wouldn't involve the ledge part, so I can't do that. I realize (assuming this is correct) that I can't use 175 as dx or 80 as dy as the projectile would go farther and lower if the cliff ledge was not there, so I'm unsure of how to solve for that. Looking for steps in the right direction. Thanks!

 

[gneill]

Hi Alex, Welcome to Physics Forums!

Start with the basic projectile motion equations for the vertical and horizontal components. Eliminate time from the vertical motion equation using the horizontal equation. You should be able to see the way forward from there.