1) Vertical direction is up, horizontal is towards the firing direction, origin is at the starting point.
2) Write two equations of motion:
a) Horizontal:
x=v_{0}cos(30)*t;
b) Vertical:
y=v_{0}sin(30)*t-gt^{2}/2; minus sign here shows that g is directed downward.
Equate y to zero and solve for t:
v_{0}sin(30)*t-gt^{2}/2=0; t=0 or t=2v_{0}sin(30)/g;
t=0 correspond to initial moment, t=2v_{0}sin(30)/g correspond to the final moment, when the shell hits the ground. This is the time of flight. Calculation yields t=35 s (g=10 m/s^{2} was used).^{
}
The time of flying up to the maximum altitude equals to the time of descent, so it took 35/2=17.5 s for the shell to reach maximum height. Plug in t=17.5 into the vertical equation of motion to get:
h=350*sin(30)*17.5-10*17.5^{2}/2≈1531 m; This is the maximum altitude.
The range can be found from the horizontal equation of motion by plugging in the total time of flight, 35 s, into the equation.
L=v_{0}cos(30)*t=350*cos(30)*35≈10609 m=10.609 km.
Answers:
a) t=35 s
b) H=1531 m.
c) L=10609 m.
It is worth mentioning that in reality the range will be just a few kilometers for such a shell, due to air resistance.
Comments