If the first meter inclined 90°, the other 89°, and so on, the total height will be:
"H=d\\text{ sin}90^\\circ+d\\text{ sin}89^\\circ+...+d\\text{ sin}23^\\circ+...+d\\text{ sin}1^\\circ=\n\\\\=d\\sum^{90^\\circ}_{i=1}\\text{sin}n_i=57.79\\text{ m}."
Using energy conservation, the final velocity:
"v=\\sqrt{2gH}=\\sqrt{2\\cdot9.8\\cdot57.79}=33.7\\text{ m\/s}."Ignoring friction and jumps at the edges where inclines are joined, we can try to calculate the time if that mass is falling from height H:
"t=\\sqrt{\\frac{2H}{g}}=\\sqrt{\\frac{2\\cdot57.79}{9.8}}=3.43\\text{ s},"
and compare it with real time calculated according to the expression
"t=\\sum^{90}_{i=1}t_i,\\\\\n\\space\\\\\nt_i=\\frac{v_i+v_{i-1}}{2d}=\\frac{v_i+v_{i-1}}{2d},\\\\\n\\space\\\\\nv_i=\\sqrt{2gd\\text{ sin}(90-i)^\\circ}.\\\\\nt=33.7\\text{ s}."
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