Question

discuss the condition for the maximum range of the projectile.Show that for a given velocity of projection, there are two angles of projection which give the same range of projectile ?

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 45⁰

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.