Answer :

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

Now the bullet 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.

The projectile motion of bullet has been shown in the figure

Now for an particle projected at an angle 𝜽 with horizontal plane or ground at an initial velocity u the maximum displacement of the particle in horizontal plane or x direction also called the range of projectile is given by

Where R is the range of the particle, u is the initial velocity of the particle, 𝜽 is the angle of projection of the particle and g denotes acceleration due to gravity

g = 9.8 ms^{-2}

Now in the case of bullet we are given

Range of the particle,

R = 3 Km = 3000m

Angle of projection,

𝜽 = 30^{o}

Let u the initial velocity of the particle

u = ?

putting the values in the equation

we get,

Or putting

Or

Now we have to adjust the gun so that bullet can go upto 5 Km or we can say range is increased to 5 Km

We can only change the angle of projection and its initial velocity u will be same or value of u^{2} will be same

Now let tange of projectile be R’

R’ = 5Km = 5000m

We have u^{2} =

We have to find angle of projection,

𝜽 = ?

Putting these values in equation

We get

Or,

But sin2𝜽 = 1.44 is not possible as for any value of 𝜽 maximum value of sine function is 1 i.e. the equation is invalid so the range R’ cannot be 5 Km so bullet cannot hit a target at a distance of 5 Km from it

The maximum range of particle is when 𝜽 = 45^{0} when

sin2𝜽 = sin 90^{o} =1

and maximum range is

Putting value of square of initial velocity of the particle

u^{2} =

we get Range

So the bullet can maximum hit a target 3.46 Km away

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