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Home » \[ s = 900 \text meters \] - Parker Core Knowledge

\[ s = 900 \text meters \] - Parker Core Knowledge

📅 May 3, 2026 👤 scraface

May 03, 2026

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After sliding: base = 7 m
New height = \( \sqrt{13^2 - 7^2} = \sqrt{169 - 49} = \sqrt{120} = 2\s...
But \( 2\sqrt{30} \approx 10.95 \), so distance slid down:
- 2\sqrt{30} \approx 12 - 10.95 = 1.05
Exact:
\Delta h = 12 - \sqrt{120} = 12 - 2\sqrt{30}
But compute exactly:
\sqrt{120} = \sqrt{4 \cdot 30} = 2\sqrt{30}
So decrease in height is \( 12 - 2\sqrt{30} \) meters.
However, the problem asks for how far the top slides down — which is t...

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