Answer:
Approximately [tex](-61)\; {\rm m\cdot s^{-2}}[/tex], assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
Under the assumptions, the vehicle would be in a free fall. Acceleration would be constant: [tex]a = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex].
Let [tex]u[/tex] denote the initial velocity of the vehicle. Let [tex]v[/tex] denote the velocity of the vehicle at the bottom of the drop. Let [tex]x[/tex] denote the displacement of the vehicle during the drop.
It is given that the initial velocity is [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex]. During the drop, displacement would be [tex]x = (-190)\; {\rm m}[/tex] (negative since the vehicle is below where it started.) The value of final velocity [tex]v[/tex] needs to be found.
It is known that the vehicle is moving downwards at the end of the fall. Therefore, the value of [tex]v\![/tex] would be negative. Apply the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex] to find [tex]v[/tex] from [tex]u[/tex], [tex]a[/tex], and [tex]x[/tex].
[tex]\begin{aligned}v^{2} &= u^{2} + 2\, a\, x\end{aligned}[/tex].
[tex]\begin{aligned}v &= -\sqrt{u^{2} + 2\, a\, x} \\ &= -\sqrt{(0\; {\rm m\cdot s^{-1}})^{2} + 2\, (-9.81\; {\rm m\cdot s^{-2}})\, (-190\; {\rm m})} \\ &\approx (-61)\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
(Note that [tex]v[/tex] is negative.)
In other words, the velocity of the vehicle would be approximately [tex](-61)\; {\rm m\cdot s^{-1}}[/tex] at the end of the drop.
