Asked by Ben
A missile is fired with a launch velocity of 15000 ft/s at a target 1200 miles away. At what angle must it be fired to hit the target? Use g = 32 ft/s.s
Answers
Answered by
bobpursley
time in air:
distance=horizveloc*t
1200=15000cosTheta*t
t=12/150costheta
in the vertical
hf=hi+Vi*t-4.8 t^2
0=0+15000sinTheta*t-4.8t^2
0=t(15000sinTehta-4.8(12/150costheta)
solutions
t=0, or
(15000sinTheta-4.8(12/150costheta)=0
15000 sinTheta*cosTheta=4.8*12/150
from trig sin2Theta=2cosT*sint
7500 sin2Theta=4.8*12/150
solve for 2*Theta, then theta.
check carefully my work
distance=horizveloc*t
1200=15000cosTheta*t
t=12/150costheta
in the vertical
hf=hi+Vi*t-4.8 t^2
0=0+15000sinTheta*t-4.8t^2
0=t(15000sinTehta-4.8(12/150costheta)
solutions
t=0, or
(15000sinTheta-4.8(12/150costheta)=0
15000 sinTheta*cosTheta=4.8*12/150
from trig sin2Theta=2cosT*sint
7500 sin2Theta=4.8*12/150
solve for 2*Theta, then theta.
check carefully my work
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