Answer :

Suppose in an projectile motion an particle is thrown making some angle 𝜽o with the horizontal with initial velocity vo sow components of the initial velocity in horizontal and vertical direction are given as


v0x = vcos and v0y = 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 vy and x component of velocity of particle be vx.


As shown in figure



So now angle of resultant velocity v at any instant with horizontal component of velocity vx is (t)


So


Now since x component of velocity constant and equal to v0x


So we have vx = v0x


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 = vy


initial velocity of particle in y direction


u = voy


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


vy = voy + (-g)t


or vy = voy – gt


so putting values of vy = voy – gt and vx = v0x


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 𝜽o with the horizontal distance covered by the projectile i.e. its range be R and the maximum height achieved by projectile be hm


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, vo is the in initial velocity of the particle, 𝜽o 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 hm is the maximum height achieved by the particle, 𝜽o 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 vo 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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