Apackage of supplies is to be dropped from an airplane so that it hits the ground at a designated spot near some campers. the airplane, moving horizontally at a constant velocity of 130 km/h, approaches the spot at an altitude of 0.520 km above level ground. having the designated point in sight, the pilot prepares to drop the package. (a) what should the angle be between the horizontal and the pilot's line of sight to the drop point when the package is released? (neglect air resistance)

Refer to the diagram below.

Note that
Air resistance is ignored.
g = 9.8 m/s²
1 km = 1000 m
1 km/h = 0.2778 m/s

The horizontal velocity is
u = 130 km/h = 130*0.2778 = 36.111 m/s
The height is
h = 0.520 km = 520 m

Let the angle of sight be x.
Then tan(x) = h/d
d = hcot(x) = 520 cot(x) m

The time of flight is
t = d/u 
  = [520 cot(x)]/36.111 
  = 14.4 cot(x)  s

Therefore, for the vertical drop,
(1/2)*g*t² = h
0.5*9.8*[14.4 cot(x)]² = 520
1016.064 cot²x = 520
cot(x) = 0.7154
x = cot⁻¹ 0.7154 = 54.4°

The angle is 54.4°  (nearest tenth)

Idon't know it i just need my points lol