A projectile is launched from the top of a tower with speed u at an angle θ above the horizontal. It crosses the horizontal level of the point of projection again after time t₁ and strikes the ground after time t₂. If t₂ = 2t₁, and H is the maximum height reached above the point of projection, then the tower height h satisfies:
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Question

A projectile is launched from the top of a tower with speed u at an angle θ above the horizontal. It crosses the horizontal level of the point of projection again after time t₁ and strikes the ground after time t₂. If t₂ = 2t₁, and H is the maximum height reached above the point of projection, then the tower height h satisfies:

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G1 Physics · Accepted Answer

The time to return to the level of projection is t₁ = 2u sinθ/g. Given t₂ = 2t₁, t₂ = 4u sinθ/g. The vertical displacement at t₂ is -h: -h = u sinθ·t₂ - ½g t₂². Substitute t₂: -h = u sinθ(4u sinθ/g) - ½g(16u²sin²θ/g²) = 4u²sin²θ/g - 8u²sin²θ/g = -4u²sin²θ/g. Hence, h = 4u²sin²θ/g. The maximum rise above the launch point is H = u²sin²θ/(2g). Therefore, h/H = [4u²sin²θ/g] / [u²sin²θ/(2g)] = 8. So, h = 8H.