A fighter plane flying horizontally at an altitude of 1.5 km with a speed of 720km/h passes directly overhead an anti-aircraft gun. What angle from the vertical should the gun be fired for the shell with muzzle speed of 600m/s to hit the plane? Also at what minimum altitude should the pilot fly the plane to avoid the hit. (Take g=10 metre per second square).... Please answer fast. Its urgently required....
Answers
Answered by
151
Given, Height = 1.5 km = 1500 m
Speed v = 720 km/h = 200 m/s
Let theta be the angle.
the velocity of the gun be fired, u = 600 m/s
Time taken = t
Horizontal distance travelled = uxt
Distance travelled by the plane = vt
Given, The shell hits the plane uxt = vt
usin theta = v
sin theta = v/u
sin theta = 200/600
= 1/3
theta = 19.5.
Hope this helps!
Speed v = 720 km/h = 200 m/s
Let theta be the angle.
the velocity of the gun be fired, u = 600 m/s
Time taken = t
Horizontal distance travelled = uxt
Distance travelled by the plane = vt
Given, The shell hits the plane uxt = vt
usin theta = v
sin theta = v/u
sin theta = 200/600
= 1/3
theta = 19.5.
Hope this helps!
Lekhraj:
Hello
Still you r left with one question..
I had asked two questions in that
Alright, the answer to the remaining question is he should fly at an altitude higher than the maximum height achieved by the shell to avoid the hit.
That only at what height?????
Wait for 2 minutes, I am solving it
I have asked clearly!!!
How will u send it to me??
Chill. I haven't seen the 2nd question properly. The answer for 2nd one is. u^2sin(90-theta)/2g
600 * 600 cos^2 theta / 2g
360000 cos^2 19.5 / 2 * 10
16004.3
16Km
600 * 600 cos^2 theta / 2g
360000 cos^2 19.5 / 2 * 10
16004.3
16Km
Answered by
43
Given, Height = 1.5 km = 1500 m
Speed v = 720 km/h = 200 m/s
Let theta be the angle.
the velocity of the gun be fired, u = 600 m/s
Time taken = t
Horizontal distance travelled = uxt
Distance travelled by the plane = vt
Given, The shell hits the plane uxt = vt
usin theta = v
sin theta = v/u
sin theta = 200/600
= 1/3
theta = sin-(1/3)
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