show that for two complementary angles of projection of a projectile thrown with thw same velocity. the horizontal ranges are equal

Solution

The maximum range is given by,

R = (u^{2} 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 45^{o}

Now,

R = (u^{2} sin2θ/g)

=> R = [u^{2} sin(180 - 2θ)/g]

=> R = [u^{2} sin{2(90 - θ)}/g]

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

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