A gun can fire a shell with a fixed initial speed u . It is aimed at a target located at a horizontal distance R on the same level. There are two different angles of projection that allow the shell to hit the target. If the difference between the times of flight for these two trajectories is t , which of the following gives the correct expression for R ?

Step-by-step solution

For a given horizontal range R and initial speed u , the two possible angles of projection are complementary: and 90^ - . The times of flight for these two angles are: t₁ = 2u g t₂ = 2u (90^ - ) g = 2u g The square of the difference in times of flight is: ( t)^2 = (t₁ - t₂)^2 = t₁^2 + t₂^2 - 2t₁ t₂ We know that: t₁^2 + t₂^2 = 4u^2 ^2 g^2 + 4u^2 ^2 g^2 = 4u^2 g^2 And the product of the times of flight is: t₁ t₂ = 4u^2 g^2 = 2 g ( u^2 2 g ) = 2R g Substituting these into the equation for ( t)^2 : ( t)^2 = 4u^2 g^2 -