Answer:
Approximately [tex]14\; {\rm m\cdot s^{-1}}[/tex].
Explanation:
The gravitational potential energy [tex]\text{GPE}[/tex] of this roller coaster is proportional to the height [tex]h[/tex] of the roller coaster.
The kinetic energy [tex]\text{KE}[/tex] of this roller coaster is proportional to the square of speed [tex]v[/tex].
The question states that the track is frictionless. Thus, during the descent, the [tex]\text{GPE}[/tex] of this roller coaster is turned into [tex]\text{KE}[/tex] without any energy loss.
When the roller coaster was at [tex](1/2)[/tex] of the the initial height, only[tex]1 - (1/2) = (1/2)[/tex] of the original [tex]\text{GPE}[/tex] was turned into [tex]\text{KE}[/tex]. The [tex]\text{KE}\![/tex] of this roller coaster at that height would be [tex]1 - (1/2) = (1/2)\![/tex] of the [tex]\text{KE}\!\![/tex] when the roller coaster is at the ground level.
The [tex]\text{KE}[/tex] of the roller coaster is proportional to [tex]v^{2}[/tex] (the square of speed [tex]v[/tex].) Thus, since the [tex]\text{KE}\![/tex] at [tex](1/2)[/tex] the initial height is [tex]1 - (1/2) = (1/2)\![/tex] the [tex]\text{KE}\!\![/tex] at the ground level, the [tex]v^{2}[/tex] at [tex](1/2)\![/tex] the initial height would also be [tex]1 - (1/2) = (1/2)\![/tex] the [tex]v^{2}\![/tex] at the ground level.
Since [tex]v = 20\; {\rm m\cdot s^{-1}}[/tex] at the ground level, [tex]v^{2} = (20\; {\rm m\cdot s^{-1}})^{2}[/tex] at the ground level. The [tex]v^{2}[/tex] at [tex](1/2)[/tex] the initial height would then be:
[tex](1 - (1/2))\times (20\; {\rm m\cdot s^{-1}})^{2}[/tex].
Thus, the speed [tex]v[/tex] at [tex](1/2)[/tex] the initial height would be:
[tex]\begin{aligned}& \sqrt{(1 - (1/2))\times (20\; {\rm m\cdot s^{-1}})^{2}} \\ =\; & \sqrt{\frac{1}{2} \times (20\; {\rm m\cdot s^{-1}})^{2}} \\ =\; & \sqrt{200}\; {\rm m\cdot s^{-1}} \\ \approx\; & 14\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
