well for part a we only need to do the vertical problem
Vi = 1690*sin55.7 = 1396 m/s straight up
v = Vi - 9.8 t
v = 0 at top
9.8 t = 1396 at top
t = 142 seconds upward so 2*142.5 in air
t in air = 285 seconds = 4.75 min PART A
u = horizontal speed forever = 1690cos 55.7 = 952 m/s
.952 km/s *285 seconds = 271 km
An artillery shell is fired at an angle of 55.7◦ above the horizontal ground with an
initial speed of 1690 m/s. The acceleration of gravity is 9.8 m/s^2.
a.Find the total time of flight of the shell, neglecting air resistance.
Answer in units of min.
b.Find its horizontal range, neglecting air resistance.
Answer in units of km
1 answer
1 answer