How do I solve projectile motion

Question

An archer on ground that has a constant upward slope of 30 degrees aims at a target 60.0 m farther up the incline. The arrow in the bow and the bull's eye at the center of the target are each 1.50 m above ground. The initial velocity of the arrow just after it leaves the bow has magnitude 32.0 m/s. At what angle above the horizontal should the archer aim to hit the bull's-eye?

If there are two such angles, calculate the smaller of the two. You might have to solve the equation for the angle by angle iteration, that is, by trial and error.

How does the angle compare to that required when the ground is level, with zero slope?

b)repeat above for ground that has downward slope of 30 degrees.

I don't want the answer (even if this is yahoo answers)

I need to figure out how to do it.

So if someone can help me get started. I'm so lost on physics it isn't even funny.

Thanks!

Answer 1

The eq for the flight path is

y = [-g(sec²Θ)/(2Vo²)]*x² + x*tanΘ

In your case, x = 60cos30°, y = 60sin30°, Vo = 32 and g = 9.8

Using these values in the eq above, you should be able to iterate for a value of Θ.

b)  Same procedure with y = -60sin30°........

Good luck.

Answer 2

Make a rough sketch of the situation. Distances and angles.