A gun 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°

Answer: (74.8°, 60.2°)