Finding the Optimal Angle: Solving the Archer Projectile Problem

[sarabeara1990]

Homework Statement

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.

Homework Equations

|V₀|= sqrt(V_x² + V_y²)
arctan(V_y/V_x) = theta
phi = theta + 30 degrees
V_x = 32sin(theta)
V_y = 32cos(theta)
V_y = V_0y-gt
X = V_0xt
y = V_0yt

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 = V_0xt in order to find my x velocity component. I also plugged it into V_y = -gt to find my y component. I then took the arctan of the y over x.

 

[rl.bhat]

Let φ be the angle between velocity and the inclined plane.
Turn the axis so that let the x-axis be along the inclined plane and the y-axis is perpendicular to it.
Take the component of g along these two axis. Now
x = (vx)*t - 1/2*(gx)*t^2...(1)
Since arrow and the the bull's eye are in the same level, net displacement (y) is zero. So
y = (vy)t - 1/2*(gy)*t^2 ...(2)
From the second equation find t. Substitute this value of t in equation (1) and solve for φ.