# 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?

Parth Shrivastava
34 Points
11 years ago

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

G Ashwin -
24 Points
10 years ago

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.

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