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a fighter plane flying horizontally at an altitude of 1.5 km with speed 720km/hr passes directly overhead an antiaircraft gun.At what angle from the vertical should the gun be fired for the shell with muzzle speed 600m/s to hit the plane?

8 years ago

Answers : (3)

Parth Shrivastava
34 Points
							

Height of the fighter plane = 1.5 km = 1500 m

Speed of the fighter plane, = 720 km/h = 200 m/s

Let θ be the angle with the vertical so that the shell hits the plane.

Muzzle velocity of the gun, u = 600 m/s

Time taken by the shell to hit the plane = t

Horizontal distance travelled by the shell = uxt

Distance travelled by the plane = vt

The shell hits the plane. Hence, these two distances must be equal.

uxt = vt

 

usinΘ = v

sinΘ = v/u

sinΘ = 200/600

       = 1/3

Θ = sin-1(0.33)

Θ = 19.5 degrees

8 years ago
G Ashwin -
24 Points
							

Height of the fighter plane = 1.5 km = 1500 m

Speed of the fighter plane, v = 720 km/h = 200 m/s

Let θ be the angle with the vertical so that the shell hits the plane. The situation is shown in the given figure.

Muzzle velocity of the gun, u = 600 m/s

Time taken by the shell to hit the plane = t

Horizontal distance travelled by the shell = uxt

Distance travelled by the plane = vt

The shell hits the plane. Hence, these two distances must be equal.

uxt = vt

In order to avoid being hit by the shell, the pilot must fly the plane at an altitude (H) higher than the maximum height achieved by the shell.

7 years ago
Jay Patil
24 Points
							
1st answer is correct but I feel that it's cos⊙ instead of sin . The angle being made by horizontal ground is cos
3 years ago
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