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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?`
7 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.
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
```
7 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.  ```
7 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
```
3 years ago
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