Question

A machine gun fires projectiles with an initial speed 650 m/s . At what angles will the projectiles hit a target placed at a horizontal distance of 450m and at height of 18m?

Answer 1

vertical problem:
Hi = 0
Vi = 650 sin s m/s
a = -9.81 m/s^2
v = Vi + a t = 650 sin s - 9.81 t
h = Hi + Vi t + (1/2) a t^2 = 650 sin s * t - 4.9 t^2
18 = 650 sin s * t - 4.9 t^2
horizontal problem:
u = 650 cos s
450 = 650 cos s * t
t = 450/650 cos s = .692 cos s
so
18 = 650 sin s * .692 cos s - 4.9 .692^2 cos^2 s
18 = 450 sin s cos s - 2.35 cos^2 s
18 = 225 (2 sin s cos s) - 2.35 cos^2 s
18 = 225 sin 2 s - 2.35 cos^2 s
1 = 12.5 sin 2 s - .13 cos^2 s
approx
18 = 225 sin 2 s
sin 2 s = .08
2 s = 4.6 degrees
s = 2.3 deg
Other solution shoot just about straight up
m g h = (1/2) m (650)^2
9.81 h = 211250
h = 21,534 meters
how long to top?
0 = Vi - 9.81 t
0 = 650 - 9.81 t
t = 66.3 s
2 t = 132.5 s in air
so 450 = u * 132.5
u = 3.4 m/s
3.4 = 450 cos s
s = 89.56 degrees