Q. 303.5( 18 Votes )

A fighter plane f

Answer :

Here the plane will move forward in horizontal direction in some time t and shell will cover the same horizontal distance along with reach the altitude or height of the plane

Now this is the case of projectile motion, the motion of Shell will be a projectile and its path will be parabola, the motion of shell can be divided into horizontal component and vertical component i.e. motion along x axis and motion along y axis

Initial speed of shell is u = 600 m/s

And let it be making an angle 𝜽 with the vertical

So horizontal component of velocity

ux = u × sin𝜽 = 600sin𝜽

and vertical component of velocity

uy = u × cos𝜽 = 600cos𝜽

Here the velocity of fighter plane is uniform and is in horizontal direction let it be

v1 = 720 km/h

converting it to m/s

1 Km = 1000m and 1 hour = 60 × 60 = 3600 s

So v1 = 72 × 1000m/3600s = 720 × 5/18 = 200 m/s

Let us consider in time t sec plane moves a distance X with uniform speed 20 m/s is horizontal direction in the same time the shell moved same distance X in same time t in horizontal direction due to horizontal component of velocity and moved a distance of Y = 1.5 Km = 1500 m equal to altitude of plane in vertical direction in order to hit the plane

The situation has been shown in the figure

Now the shell is only experiencing one force in vertically downward direction because of its weight (gravitational force of earth) due to which its accelerating along negative y axis due to acceleration due to gravity, acceleration is uniform i.e. all the three equations of motion are valid in along Y direction, now there is no force in horizontal direction so component of velocity in horizontal direction or along X axis would remain constant.

Since shell and plane cover equal distance in equal time in horizontal direction with uniform speed so horizontal component of velocity of shell and speed of plane must be equal

i.e ux = v1

or u × sin𝜽 = 600 × sin𝜽 = 200 m/s

or sin𝜽 = 20/600 = 1/3

now so we get

so the gun must be making an angle of 19.47o with the vertical

In order to avoid being hit by the gun the plane must fly above the the maximum vertical height shell can reach, the motion of shell is a projectile as explained above and for the maximum height of projectile we will apply equation of motion

in y direction

Where v is the final velocity, u is the initial velocity of the particle, a is the acceleration and s is the displacement of the particle

For y direction we have

Where Initial velocity of particle in vertical direction making an angle 𝜽 with the vertical

uy = u × cos𝜽 = 600cos𝜽

since upon reaching maximum height Shell will come to rest to final velocity in vertical direction

vy = 0 m/s

displacement of shell in y direction must be equal to height of plane let the maximum height of plane be

Sy = Y = ?

The acceleration of particle is in downward direction and is equal to acceleration due to gravity

a �y = -g = -10 ms-2

putting values in equation)

02 – (600ms-1× cos𝜽)2 = 2 × (-10 ms-2) ×Y

So we have

Here cos2𝜽 = cos2(19.47o)

We know sin2(19.5o) = (1/3)2=1/9

Using cos2𝜽 + sin2𝜽 = 1

cos2(19.47o) = 1 - sin2(19.5o) = 1 – 1/9 =8/9

so putting value, we know maximum height of shell can be

So the fighter plane must fly above 16 Km from ground in order to avoid being hit by shell

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