Answer :

Suppose in an projectile motion an particle is thrown making some angle 𝜽_{o} with the horizontal with initial velocity v_{o} sow components of the initial velocity in horizontal and vertical direction are given as

v_{0x} = vcos and v_{0y} = vsin𝜽

there is no horizontal force on the particle so horizontal component of velocity will remain constant and due to acceleration due to gravity the motion in y direction will be uniformly accelerated where y component of velocity will change and all equations of motion will be valid in y direction

A. Now at any instant of projectile motion let the angle of velocity of particle with x axis or horizontal be 𝜽(t) now 𝜽(t) is varying because y component of velocity of particle is constantly changing with time so the angle of resultant velocity is also changing with time , now angle of resultant velocity of the particle or the velocity of particle with x axis at any instant is same as angle with x component or horizontal component of velocity at any instant let the y component of velocity of particle be v_{y} and x component of velocity of particle be v_{x}.

As shown in figure

So now angle of resultant velocity v at any instant with horizontal component of velocity v_{x} is (t)

So

Now since x component of velocity constant and equal to v_{0x}

So we have v_{x} = v_{0x}

Now to find y component of velocity at any instant we will aplly equation of motion

v = u + at in y direction

here v is the final velocity, u is the initial velocity t is the time and a is the acceleration of particle

for y direction velocity of particle in y direction is

v = v_{y}

initial velocity of particle in y direction

u = v_{oy}

particle is accelerating because of acceleration due to gravity is in vertically downward direction or negative y direction so

a = -g (g is acceleration due to gravity)

so y component of velocity of particle at any time t sec is

v_{y} = v_{oy} + (-g)t

or v_{y} = v_{oy} – gt

so putting values of v_{y} = v_{oy} – gt and v_{x} = v_{0x}

we have

hence angle between the velocity and x axis at any instant is given by

B. Now let the projectile be launched at initial angle 𝜽_{o} with the horizontal distance covered by the projectile i.e. its range be R and the maximum height achieved by projectile be h_{m}

as shown in the figure below

Now for an projectile the maximum horizontal distance or the range is given by the equation

Where R is the range, v_{o} is the in initial velocity of the particle, 𝜽_{o} is the angle of projection and g is the acceleration due to gravity

Now the maximum height of the projectile in vertical direction is given by the equation

Where h_{m} is the maximum height achieved by the particle, 𝜽_{o} is the angle of projection and g is the acceleration due to gravity

Now diving these equations, we get

Or

Since sin2𝜽 = 2sin𝜽cos𝜽

And cancelling out g and v_{o} we get

Or we have

So we get

So

i.e. the initial angle of projection for an projectile can be given as

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