From the top of a tower of height h, two particles are projected with the same speed u. One is projected at angle +θ and the other at angle -θ with the horizontal.if their horizontal ranges on the ground are R₊ and R₋, then R₊R₋ equals:
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Question

From the top of a tower of height h, two particles are projected with the same speed u. One is projected at angle +θ and the other at angle -θ with the horizontal.if their horizontal ranges on the ground are R₊ and R₋, then R₊R₋ equals:

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G1 Physics · Accepted Answer

Let the corresponding times of flight be T₊ and T₋. For +θ: -h = u sinθ·T₊ - ½gT₊². For -θ: -h = -u sinθ·T₋ - ½gT₋². Thus, T₊ = [u sinθ + √(u²sin²θ+2gh)]/g, T₋ = [-u sinθ + √(u²sin²θ+2gh)]/g. Their product is T₊T₋ = [(u²sin²θ+2gh) - u²sin²θ]/g² = 2h/g. Now, R₊ = u cosθ·T₊, R₋ = u cosθ·T₋. Hence, R₊R₋ = u²cos²θ·T₊T₋ = 2hu²cos²θ/g.