Answer:
Approximately [tex]50\; \rm m \cdot s^{-2}[/tex], assuming that air resistance is negligible and that the rock was released with zero initial velocity.
Step-by-step explanation:
Let [tex]g[/tex] denote the gravitational acceleration on this planet.
Let [tex]x[/tex] denote the displacement of this rock during this fall. Let [tex]t[/tex] denote the duration of this fall. Assume that the rock was released with an initial velocity [tex]v_{0} = 0\; \rm m \cdot s^{-1}[/tex].
Assume that the air resistance is negligible (such that gravity is the only force on this rock). The acceleration [tex]a[/tex] of the rock during the fall would be constantly equal to the gravitational acceleration [tex]g[/tex].
Since the acceleration of the rock during the fall was constantly [tex]a = g[/tex], the following SUVAT equation would apply:
[tex]\displaystyle x = \frac{1}{2}\, a \, t^{2} + v_{0}\, t[/tex].
Rearrange this equation to find an expression for acceleration [tex]a[/tex]:
[tex]\begin{aligned}a &= \frac{x - v_{0}\, t}{t^{2} / 2}\end{aligned}[/tex].
Convert the height and duration of this fall to standard units:
[tex]\begin{aligned}h &= 89\; \rm km \times \frac{10^{3}\; \rm m}{1\; \rm km} \\ &= 8.9\times 10^{4}\; \rm m\end{aligned}[/tex].
[tex]\begin{aligned}t &= 0.99\; \rm min \times \frac{60\; \rm s}{1\; \rm min} \\ &= 59.4\; \rm s\end{aligned}[/tex].
Substitute in the values to find the value of acceleration [tex]a[/tex]:
[tex]\begin{aligned}a &= \frac{x - v_{0}\, t}{t^{2} / 2} \\ &= \frac{89000\; \rm m}{(59.4\; \rm s)^{2} / 2} && (\text{Note that $v_{0} = 0\; \rm m\cdot s^{-1}$.}) \\ &\approx 50\; \rm m\cdot s^{-2} \end{aligned}[/tex].
By the assumptions above, the acceleration of this rock during this fall would be equal to the gravitational acceleration on this planet. Thus, [tex]g \approx 50\; \rm m\cdot s^{-2}[/tex] on this planet.
