CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

A fighter plane flying horizontally at an altitude of 1.5km with a speed of 720kmh-1 passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired the shell with muzzle speed 600ms-1 to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take g=10ms-2)


Open in App
Solution

Step 1: Given data

Height at which plane is flying, h=1.5km=1500m

Speed of plane, v=720kmh-1=720×518=200ms-1

Speed of bullet of firing gun, u=600ms-1

Acceleration due to gravity, a=10ms-2

Step 2: Calculation

Step 3: Finding the angle at which a gun should be fired for the plane to hit its target, θ

Distance traveled by plane =vt (t=time taken by the shell to hit the plane )

Distance traveled by the shell in x-direction = uxt

These 2 distances are equal since the shell hits the plane

vt=uxtv=uxv=usinθsinθ=vusinθ=200600sinθ=0.33θ=sin-1(0.33)θ=19.27

Step 4: Finding the height at which the pilot should fly the plane to avoid getting hit.

The pilot should fly the plane at a height more than the maximum height of launch of the shell at any angle

According to projectile motion .Hmax=u2sin2(90o-θ)2g=600×600×0.892×10=16.02km

The angle at which the shell should be launched is 19.27and the height at which the plane should be kept to avoid the shell attack should be more than 16.02km.


flag
Suggest Corrections
thumbs-up
0
BNAT
mid-banner-image
similar_icon
Similar questions
View More