Agun has muzzle speed 144 m/s. find two angles of elevation that can be used to hit a target 780 m away. (enter your answers as a comma-separated list. round your answers to one decimal place. use g â 9.8 m/s2.)

The muzzle speed is  144 m/s.
Let x = the angle of elevation.
Then
u = 144 cos(x), the horizontal velocity
v = 144 sin(x), the vertical launch velocity

Assume no air resistance, and g = 9.8 m/s².

To hit the target 780 m away, the time of flight is
t = (780 m)/[144 cos(x)] = 5.4167 sec(x) s

In terms of the vertical velocity,
(144 sin(x) m/s)*(5.4167 sec(x) s) + (1/2)*(-9.8 m/s²)*(5.4167 sec(x) m/s)² = (780  m)
780 tan(x) - 143.769 sec²x = 780
tan(x) - 0.1843 (1 + tan²x) = 1
0.1843 tan²(x) - tan(x) + 1.1843 = 0
tan²x - 5.4259 tan(x) + 6.4259 = 0

Solve with the quadratic formula.
tan(x) = 0.5[5.4259 +/- √(3.737)] = 3.6795 or 1.7464
x = tan⁻¹ 3.6795 = 74.8°
or
x = tan⁻¹ 1.7464 = 60.2°

(74.8°, 60.2°)