A projectile is launched from the top of a tower of height h with speed u. Its maximum rise above the point of projection is exactly h.if u² = 4gh, and it strikes the ground with its velocity making angle φ below the horizontal, then tanφ is:
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Question

A projectile is launched from the top of a tower of height h with speed u. Its maximum rise above the point of projection is exactly h.if u² = 4gh, and it strikes the ground with its velocity making angle φ below the horizontal, then tanφ is:

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G1 Physics · Accepted Answer

Maximum rise above launch point: H = u²sin²θ/(2g). Given H=h, u²sin²θ/(2g) = h. Using u²=4gh, 4gh sin²θ/(2g) = h ⇒ 2sin²θ = 1 ⇒ θ = 45°. Horizontal impact component: vₓ = u cosθ = 2√(gh)(1/√2) = √(2gh). Vertical impact component satisfies vᵧ² = u²sin²θ + 2gh = 2gh + 2gh = 4gh. Thus, |vᵧ| = 2√(gh). Therefore, tanφ = |vᵧ|/vₓ = [2√(gh)]/[√(2gh)] = √2.