1. The problem statement, all variables and given/known data An archer on ground that has a constant upward slope of 30.0 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 the ground. The initial velocity of the arrow just after it leaves the bow has magnitude 32.0 m/s. A)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 iteration—that is, by trial and error. B)How does the angle compare to that required when the ground is level, with 0 slope? C)Repeat the above for ground that has a constant downward slope of 30.0 degrees. 2. Relevant equations |V0|= sqrt(Vx2 + Vy2) arctan(Vy/Vx) = theta phi = theta + 30 degrees Vx = 32sin(theta) Vy = 32cos(theta) Vy = V0y-gt X = V0xt y = V0yt 3. The attempt at a solution My issue is finding time. I tried just basic V=D/T but I know that isn't the correct way. Once I have time, I could find my x and y velocity components and take the arctan to find my angle theta. When using my time from V=D/T I plugged it into x = V0xt in order to find my x velocity component. I also plugged it into Vy = -gt to find my y component. I then took the arctan of the y over x.