Question

show the for two complementary angle of projection of a projectile with velocity the horizontal ranges are equal..​

Answer 1

The maximum range is given by,

R = (u2 sin2θ/g)

[u is the initial velocity of the projectile which is projected at an angle θ to the horizontal]

The range is maximum when sin2θ is maximum.

Therefore, sin2θ = 1

=> 2θ = π/2

=> θ = π/4 or 45o

Now,

R = (u2 sin2θ/g)

=> R = [u2 sin(180 - 2θ)/g]

=> R = [u2 sin{2(90 - θ)}/g]

So, for two angles, θ and 90 – θ the range of the projectile will be same.

Answer 2

Answer:

The maximum range is given by,

R = (u2 sin2θ/g)

[u is the initial velocity of the projectile which is projected at an angle θ to the horizontal]

The range is maximum when sin2θ is maximum.

Therefore, sin2θ = 1

=> 2θ = π/2

=> θ = π/4 or 45o

Now,

R = (u2 sin2θ/g)

=> R = [u2 sin(180 - 2θ)/g]

=> R = [u2 sin{2(90 - θ)}/g]

So, for two angles, θ and 90 – θ the range of the projectile will be same