English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

For the ballistic missile aimed to achieve the maximum range of 16500 km, what is the maximum altitude reached in the trajectory?

2007-09-16 15:27:39 · 1 answers · asked by mathgeek 1 in Science & Mathematics Physics

1 answers

I can only provide an approximation using a flat-earth ballistic model (ref.). The exact solution would require orbital mechanics. Assuming a=9.8 m/s^2, launch angle theta = 45 deg (maximum-range angle).
v0 = sqrt(range*a/sin(2theta)) = 12716.1314872095 m/s
v0x = v0y = sin(theta)*v0 = 8991.66280504265 m/s
time = range/v0x = 1835.03322551891 s
ymax = v0y^2/(2a) = 4125000.00002105 m
For a 45 deg launch, max altitude always = range/4. This is because range = v0^2sin(2theta)/g and ymax = v0^2sin^2(theta)/(2g), and (trigonometric identity) sin(2theta) = 2sin(theta)cos(theta) = (at 45 deg) 2sin^2(theta). Substituting, range = v0^2*2sin^2(theta)/g = 4*ymax.

2007-09-18 04:19:23 · answer #1 · answered by kirchwey 7 · 0 0

fedest.com, questions and answers