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Using conservation of energy, how fast would a rollercoaster be traveling at a point half way above the ground where it started from, assuming that is started at rest and 85.3 m above the ground?

User Nikkie
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\begin{gathered} ME=KE+PE \\ ME\text{ conserved} \\ \end{gathered}
\begin{gathered} KE1+PE1=KE2+PE2 \\ v1=0 \\ h1=85.3m \\ v2=? \\ h2=42.65m \end{gathered}
\begin{gathered} (mv1^2)/(2)+mgh1=(mv2^2)/(2)+mgh2 \\ 0+mgh1=(mv2^2)/(2)+mgh2 \\ g(h1-h2)\cdot2=v2^2 \\ 9.81\cdot(42.65)\cdot2=v2^2=836.793 \\ 28.93m/s=v2 \end{gathered}

First, the energy is conserved so the energy in point 1 is the same as the energy at point 2

Then you have two types of energy, potential energy and, kinetic energy

With the values, you replace in the formula and solve the problem

User Saqib Naseeb
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