Question

How do I go about solving a non-horizontally launched projectile problem ("launched [x] m/s [x] degrees above the horizontal") if it includes a starting height (cliff)?

Answer 1

You will have equations for both the x (horizontal) and y (vertical) motion. For each direction you need the initial velocity, the acceleration and the initial position. You get the x-component of the initial velocity by using the cosine of the firing angle [Vx = V cos (angle)] and the y-component of the initial velocity using the sin of the firing angle. You can usually let the initial x position be zero. The initial y position is usually the height of the cliff. There is usually no acceleration in the x-direction unless there is wind resistance, but there will be acceleration due to gravity in the y direction.
The big deal is to figure out what quantities are positive or negative. Here's the deal: if you made the positive y direction up, make the acceleration due to gravity negative (since it is down and down is negative); if you made the positive y direction downward, make the acceleration due to gravity positive (since you defined down as positive). Careful - consider the big picture: if up is positive, and the projectile's initial velocity is down (cannon aimed downward), then you must make the initial y-velocity negative as well as making the acceleration due to gravity negative.
Work either the x or y equation until you need a "missing" variable - like time. You switch to the other dimension to get the missing variable... then plug it back in to the equation needing that info.

Answer 2

Awesome; thank you so much!