Question

An airplane with a speed of 88.2 m/s is climbing upward at an angle of 55.6 ° with respect to the horizontal. When the plane's altitude is 523 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

Answer 1

Vi = +88.2 sin 55.6 = 72.8
u = 88.2 cos 55.6 until it hits the ground = 49.8

v = Vi - 9.81 t
h = Hi + Vi t - 4.9 t^2

so when is h = 0 ?
0 = 523 + 72.8 t - 4.9 t^2
solve quadratic
t = 20.2 seconds (or -5.3 seconds but that was when it might have been thrown from the ground :)
https://www.mathsisfun.com/quadratic-equation-solver.html
x = u t = 49.8*20.2 = 1007 meters or about a kilometer
v = Vi - 9.81 (20.2) = -125 m/s
u is still 88.2
so tan angle down from level = 125/88.2
angle down from level = 54.8

Answer 2

x = u t = 49.8*20.2 = 1007 meters or about a kilometer
v = Vi - 9.81 (20.2) = -125 m/s

u is still 49.8 NOT 88.2 !!!!!!!

so tan angle down from level = 125/49.8
angle down from level = 68.3 deg