Answer:
Approximately [tex]25\; {\rm m\cdot s^{-1}}[/tex] (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] and that air resistance is negligible.)
Explanation:
Under the assumptions, the rock would accelerate at [tex]a = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex]. It is also given that the initial velocity is [tex]u = (-4)\; {\rm m\cdot s^{-1}}[/tex] and displacement [tex]x = (-30)\; {\rm m}[/tex]. Note that the value of [tex]a[/tex], [tex]u[/tex], and [tex]x[/tex] are negative because these vector quantities point downward.
Apply the following SUVAT equation to find the magnitude of velocity [tex]v[/tex] right before landing:
[tex]v^{2} - u^{2} = 2\, a\, x[/tex].
Rearrange to obtain:
[tex]\begin{aligned}v^{2} &= u^{2} + 2\, a\, x \end{aligned}[/tex].
[tex]\begin{aligned}|v| &= \sqrt{u^{2} + 2\, a\, x} \\ &= \sqrt{(-4)^{2} + 2\, (9.81)\, (30)}\\ &\approx 25\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
In other words, the speed (magnitude of velocity) of the rock would be approximately [tex]25\; {\rm m\cdot s^{-1}}[/tex] right before landing.
