Question

An enemy fighter jet is flying at a constant height of 250 m with a velocity of 500 m/s. The fighter jet passes over an anti – aircraft gun that can fire at any time and in any direction with a speed of 100 m/s.
For how much time, it will be in the danger to be hit ?​

Answer 1

Explanation:

Given An enemy fighter jet is flying at a constant height of 250 m with a velocity of 500 m/s. The fighter jet passes over an anti – aircraft gun that can fire at any time and in any direction with a speed of 100 m/s.For how much time, it will be in the danger to be hit ?

• We need to find the time. Now the jet is in danger and anytime it may be hit.
• So the velocity of the jet is 500 m/s and the height is 250 m
• We have the equation of the parabolic path that is
• So y = x tan θ x ½ gx^2 sec^2 θ / u^2 ---------1  since all are constant and θ changes we have
• So dy / d(tan θ) = x – 1/2u^2 gx^2 2 tan θ -------2
• So putting dy / d(tan θ) = 0 we get
• So x – ½u^2 g x^2 2 tan θ = 0
• So tan θ x gx^2 / u^2 = x
•      Or tan θ = u^2 / gx
• Substituting the value of tan θ in equation 1 we get
• So y = x tan θ – 1/2u^2 x gx^2 sec^2 θ (sec^2 θ = 1 + tan^2 θ)
• So y = x u^2 / gx – 1/2u^2 x gx^2 (1 + u^4 / g^2 x^2)
• So y = u^2 / g – gx^2 / 2u^2 – u^2 / 2g
• Or y = u^2 / 2g – gx^2 / 2u^2
• Or y max = u^2 / 2g – gx^2 / 2u^2
• Now y max ≥ 250
• So u^2 / 2g – gx^2 / 2u^2 ≥250
• So the velocity is 100 m/s
• So (100)^2 / 2 x 10 – 10 x^2 / 2 x (100)^2 ≥ 250
• So 10000 / 20 – 10 x^2 / 20,000 ≥ 250
• So 500 – 250 ≥ 10 x^2 / 20,000
• 250 x 2000 ≥ x^2
• Or x = ±√250 x 2000
• Or we get – 500 √2 ≤ x ≤ 500√2
• Therefore time t = x1 + x2 / 500
•                              = 500 √2 + 500√2 / 500
•              So t = 2√2 secs

Reference link will be

https://brainly.in/question/7080383