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)

Well it depends on the object or substance

Answer;

hot magma rises and pushes up toward the lithosphere, and as it cools it cycles back down toward the core to be heated again.

explanation; convection currents are the circular pathways taken by a fluid that enhances heat transfer by the process known as convection which occurs in fluids.convection currents transfer heat from one place to another by mass motion of a fluid. earth mantle is in a semi-liquid state and behaves like any other fluid, rising up from the bottom of the mantle after becoming hotter and less dense from heat of the earth's core, as the rock loses heat into the earth's crust it becomes cooler and more dense, then sinking back down to the core, then repeating the cycle.