It is possible to project a particle with a given velocity in two possible ways so as to make them pass through a point P at a horizontal distance r from the point of projection, if t1 and t2,are times taken to reach this going in two possible ways, then the product t1t2 is proportional to

Explanation

Here's how to solve this projectile motion problem and find the relationship between the product of times (t₁t₂) and the horizontal distance (r):1. Define variables and set up the equations:Let u be the initial velocity of the projectile.Let θ₁ and θ₂ be the two possible launch angles that reach point P.Let t₁ and t₂ be the times taken to reach point P for angles θ₁ and θ₂, respectively.Let g be the acceleration due to gravity.For both launch angles, we can use the following kinematic equations:Horizontal motion:x = u * t * cos(θ)where:x is the horizontal distance (which is always r in this case)t is the timeVertical motion:y = u * t * sin(θ) - 1/2 * g * t²Since the particle reaches the same horizontal distance (r) in both cases, we can write:r = u * t₁ * cos(θ₁) = u * t₂ * cos(θ₂)2. Find the relationship between times (t₁ and t₂):Since the particle reaches the same height (y = 0) at point P for both angles, we can write:0 = u * t₁ * sin(θ₁) - 1/2 * g * t₁²0 = u * t₂ * sin(θ₂) - 1/2 * g * t₂²Dividing both equations by their respective horizontal distances (r):0 = (u * sin(θ₁) / r) * t₁ - 1/2 * g * (t₁²/r²)0 = (u * sin(θ₂) / r) * t₂ - 1/2 * g * (t₂²/r²)Since we have two equations with two unknowns (θ₁ and θ₂), we can try to eliminate one of them.3. Eliminate launch angles (θ₁ and θ₂):From the horizontal distance equations, we can write:sin(θ₁) = r / (u * t₁)sin(θ₂) = r / (u * t₂)Substitute these expressions into the two vertical motion equations:0 = (r / t₁) - 1/2 * g * (t₁²/r²)0 = (r / t₂) - 1/2 * g * (t₂²/r²)Multiply both sides of both equations by 2t₁t₂:0 = 2r * t₂ - g * t₁t₂0 = 2r * t₁ - g * t₁t₂Adding these two equations:0 = 2r * (t₁ + t₂) - 2g * t₁t₂Dividing both sides by -2g:t₁t₂ = r / g to option (c).