To find the angle θ, the horizontal distance, x needs to be known. However, the package is moving in both the x and y direction.

First, start with a diagram at the instant the package is released as shown at the left. Position the origin of the Cartesian coordinate system at the point where the package is released, with the positive y axis pointing up. Resolve the acceleration into rectangular components,

and integrate these equations over time to get velocity, and then again to get position.

In addition to the two differential equations, initial conditions (i.e. boundary conditions) for time, position and velocity are needed to solve the equations. The initial conditions are:

     Initial Position:  x_o = 0, y_o = 0
     Initial Velocity:  v_xo, = 176 ft/s, v_yo = 0
     Initial Time:  t_o = 0

Integrating the acceleration equation with respect to time gives,

     dv_x = a_x dt

     v_x = v_xo = 176 ft/s

Since the x-direction velocity is constant, the horizontal velocity does not change while the package is falling.

The horizontal position can be determined by integrating v_x = over time,

     dx = v_x dt

     x = 176 t

The difficulty with this result, is the time for the package to hit the ground is not known. Thus, vertical direction now needed to determine t.

Just like for the horizontal direction, the vertical velocity and position can be determined by integrating the acceleration equation,

     dv_y = a_y dt

     v_y = -gt

Integrating v_y over time will give the y position,

     dy = v_y dt

The vertical distance to the ground is known, 600 ft,

    600 - 0 = -g/2 (t - 0)²

     t = 6.10 s

     x = 176 (6.10) = 1,074 ft

Trigonometry can be used to find the angle θ:

     θ = tan^-1 (600/1074) = 29.2^o

The falling package exemplifies the characteristics of projectile motion.