Answer:
Approximately [tex]31\; {\rm m\cdot s^{-1}}[/tex] assuming that there is no friction on the rollercoaster, and that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
Make use of the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex], where:
If there is no friction on the rollercoaster, the acceleration of the rollercoaster would be equal in magnitude to the gravitational field strength, [tex]g[/tex]:
[tex]a = g = 9.81\; {\rm m \cdot s^{-2}}[/tex].
The initial velocity of this rollercoaster is [tex]0\; {\rm m\cdot s^{-1}}[/tex] (that is, [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex]) since the rollercoaster was initially stationary. The displacement of this rollercoaster would be [tex]x = 49\; {\rm m}[/tex] (same as the height of the hill.)
Rearrange the equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex] to find an expression for the final velocity [tex]v[/tex] of this rollercoaster:
[tex]\begin{aligned}v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 9.81\; {\rm m\cdot s^{-2} \times 49\; {\rm m} + (0\; \rm m\cdot s^{-1}})^{2}}\\ &\approx 31\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
