A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km/h. At what angle of sight (w.r.t. horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target?
A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km/h. At what angle of sight (w.r.t. horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target?
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1 Answer
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This is a Short Answer Type Question as classified in NCERT Exemplar
Explanation- when it be at position P, drops a bomb to hit a target T
Let
Speed of the plane =720km/h = 720 = 200m/s
Altitude of the plane P’T = 1.5km= 1500m
If bomb hits the target after time t then horizontal distance travelled by the bomb PP’=u =200t
Vertical distance travelled by the bomb P’T=1/2gt2
1500 = ½ (9.8)t2
So t2= 1500/4.9, t =
PP’=200 (17.49)m=
tan =P’T/P’P=1500/200 (17.49)=0.49287= tan23012’
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Please find the solution below:
after 10 kicks,
v? = 3tî v? = 24cos 60°î + 24sin 60°? = 12î + 12√3?
v? = v? – v? = (12 – 3t)î + 12√3?
It is minimum when 12 - 3t = 0 ⇒ t = 4sec
ω = θ² + 2θ
α = (ωdω)/dθ = (θ² + 2θ) (2θ + 2)
At θ = 1rad.
ω = 3rad/s and α = 12rad/s²
a? = αR = 12 m/s² a? = ω²R = 9 m/s² A? = √ (a? ² + a? ²) = 15 m/s²
a? = v? ²/4r
a_A? = (v? ²/r²) × r = v? ²/r
a_A = 3v? ²/4r
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