Asked by Rahma
An aircraft gun fires at an elevation of 60 degree at an enemy aircraft ten times ten raise to power four m above the ground at what speed must the cannon be shot to hit the plane at that height [take g= nine point eight meter per second square
Answers
Answered by
Anonymous
Fires at speed S
vertical component at muzzle = Vi = s sin 60 = 0.866 s
vertical problem
v = Vi - g t = 0.866 s - 9.8 t
top when v = 0
so at top when t = 0.866 s / 9.8 = 0.0877 s
h = Vi t - 4.9 t^2
10 * 10^4 = 0.866 s * 0.0877 s - 4.9 (0.0877)^2 s^2
100,000 = 0.0759 s^2 - 0.0377 s^2 = 0.038 s^2
s = 1620 meters/ second
vertical component at muzzle = Vi = s sin 60 = 0.866 s
vertical problem
v = Vi - g t = 0.866 s - 9.8 t
top when v = 0
so at top when t = 0.866 s / 9.8 = 0.0877 s
h = Vi t - 4.9 t^2
10 * 10^4 = 0.866 s * 0.0877 s - 4.9 (0.0877)^2 s^2
100,000 = 0.0759 s^2 - 0.0377 s^2 = 0.038 s^2
s = 1620 meters/ second
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