When the coaster starts to descend, its potential energy gets converted into kinetic energy. Mathematically,
\(\frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = mgh\)
Here,\(m\)is the mass of the coaster,\(g\)is the acceleration due to gravity\(\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\),\({v_f}\)is the final velocity of the coaster,\({v_i}\)is the initial velocity of the coaster\(\left( {{v_i} = 5{\rm{ m}}/{\rm{s}}} \right)\), and h is the height\(\left( {h = 20.0{\rm{ m}}} \right)\).
As a result, the final velocity of the coaster is given as,
\({v_f} = \sqrt {v_i^2 + 2gh} \)
Putting all known values,
\(\begin{aligned}{c}{v_f} = \sqrt {{{\left( {5{\rm{ m}}/{\rm{s}}} \right)}^2} + 2 \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {20.0{\rm{ m}}} \right)} \\ = 20.42{\rm{ m}}/{\rm{s}}\end{aligned}\)
When the coaster ascends uphill, its kinetic energy gets converted into potential energy.
\(\frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = mgh'\)
Here, \(m\) is the mass of the coaster, \(g\) is the acceleration due to gravity \(\left( {g = - 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\), \({v_f}\) is the final velocity of the coaster \(\left( {{v_f} = 0} \right)\), \({v_i}\) is the initial velocity of the coaster \(\left( {{v_i} = 5{\rm{ m}}/{\rm{s}}} \right)\), and \(h'\) is the height above the starting point.
Therefore, the height above the starting point attained by the coaster is,
\(h' = \frac{{v_f^2 - v_i^2}}{{2g}}\)
Putting all known values,
\(\begin{aligned}{}h' = \frac{{{{\left( 0 \right)}^2} - {{\left( {5{\rm{ m}}/{\rm{s}}} \right)}^2}}}{{2 \times \left( { - 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)}}\\ = 1.28{\rm{ m}}\end{aligned}\)
When the coaster stops and starts to roll down, its initial velocity becomes zero\(\left( {{v_i} = 0} \right)\), and it is a height of,
\(\begin{aligned}{}H = h + h'\\ = \left( {20.0{\rm{ m}}} \right) + \left( {{\rm{1}}{\rm{.28 m}}} \right)\\ = 21.28{\rm{ m}}\end{aligned}\)
While rolling down the potential energy of the coaster gets converted into kinetic energy. Mathematically,
\(\frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = mgH\)
As a result, the final velocity of the coaster is given as,
\({v_f} = \sqrt {v_i^2 + 2gH} \)
Putting all known values,
\(\begin{aligned}{}{v_f} = \sqrt {{{\left( 0 \right)}^2} + 2 \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {21.28{\rm{ m}}} \right)} \\ = 20.42{\rm{ m}}/{\rm{s}}\end{aligned}\)
Therefore, in both cases the coaster will descend down with same speed.
