Question
Given an angle of 30 degrees, find the minimum initial speed of a cannon ball that travels in a horizontal distance of 15000 mi.
Answers
Damon
I suspect you mean meters not miles but whatever
s = speed
horizontal:
u = s cos 30 = .866 s
if t = time rising, then 2t = time in air
15,000 = u *2 t = 1.732 s t
so
s t = 8660
vertical
Vi = s sin 30 = .5 s
v = Vi - 9.81 t
v = 0 at t (the top)
0 = .5s - 9.81 t
t = .05096 s
so
s (.05096 s) = 8660
s = 412 meters/second
s = speed
horizontal:
u = s cos 30 = .866 s
if t = time rising, then 2t = time in air
15,000 = u *2 t = 1.732 s t
so
s t = 8660
vertical
Vi = s sin 30 = .5 s
v = Vi - 9.81 t
v = 0 at t (the top)
0 = .5s - 9.81 t
t = .05096 s
so
s (.05096 s) = 8660
s = 412 meters/second