An object is projected from the ground such that its horizontal range is maximum and equal to 160 m . Another object is projected with the same initial speed, but its horizontal range is half of the maximum range. There are two possible angles of projection for this second object. If H₁ and H₂ are the maximum heights attained in these two possible cases, find the product H₁ H₂ (in m ^2 ).

Step-by-step solution

The maximum horizontal range is given by R_ max = u^2 g = 160 m . The range of the second object is half of the maximum range, so R = 80 m . Using the range formula R = u^2 (2 ) g , we have: u^2 (2 ) g = 80 160 (2 ) = 80 (2 ) = 1 2 . This gives 2 = 30^ or 150^ , which means the two possible angles of projection are ₁ = 15^ and ₂ = 75^ . The maximum heights attained in these two cases are: H₁ = u^2 ^2(15^ ) 2g H₂ = u^2 ^2(75^ ) 2g = u^2 ^2(15^ ) 2g The product of the maximum heights is: H₁ H₂ = ( u^2 ^2(15^ ) 2g ) (