A particle is projected from the top of a tower of height h with speed u at an angle θ above the horizontal. It strikes the horizontal ground with its velocity making an angle 45° below the horizontal. If u² = 4gh, then tanθ is:
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Question

A particle is projected from the top of a tower of height h with speed u at an angle θ above the horizontal. It strikes the horizontal ground with its velocity making an angle 45° below the horizontal. If u² = 4gh, then tanθ is:

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G1 Physics · Accepted Answer

Horizontal velocity remains constant: vₓ = u cosθ. At impact, the velocity is 45° below the horizontal, hence |vᵧ| = vₓ. If T is the time of flight, gT - u sinθ = u cosθ, so T = u(sinθ + cosθ)/g. Vertical displacement from launch point to ground is -h: -h = u sinθ·T - ½gT². Therefore, h = ½gT² - u sinθ·T. Substituting T, h = [u²/(2g)](sinθ+cosθ)² - [u²/g]sinθ(sinθ+cosθ) = [u²/(2g)](cos²θ-sin²θ) = [u²/(2g)]cos2θ. Using u²=4gh, h = 2h cos2θ ⇒ cos2θ = 1/2. For 0°<θ<45°, 2θ = 60° ⇒ θ = 30°. H…