You can use: [tex]s=ut+\frac{1}{2}at^{2}[/tex] for the entire 'experiment'.
[tex]s=displacement[/tex]
[tex]u=initial (starting) velocity[/tex]
[tex]t=time[/tex]
[tex]a=acceleration[/tex]
Note: You are trying to find the acceleration so the weight formula can be used later.
Therefore, by rearrangement, [tex]a=\frac{2(s-ut)}{t^{2}}[/tex]
Since the rock is launched at 13ms-1, this is the initial velocity (u)
The time taken for the rock to come back from where it was launched, t=1.51 seconds
This seems to be all the information given that is relevant to this equation. But it is also known that the displacement (s) will be 0 when the rock reaches back to where it left.
Hence, [tex]a=\frac{2(0-13(1.51))}{(1.51)^{2}}[/tex]
[tex]a \approx -17.22 ms^{-2}[/tex] (note: negative acceleration means the acceleration is acting opposite to the rock launch. i.e. downwards towards the planet)
Now since the acceleration has been found, the weight of the rock can be found using:
[tex]F=ma[/tex]
F= force (Newtons) / weight of object
m = mass = 5.24kg
a = acceleration = -17.22 ms-2
Therefore, [tex]F=(5.24)(17.22)[/tex]
[tex]F\approx 90kg[/tex]