Answer:
[tex]a=6.4\ m/s^2[/tex]
Explanation:
Given that,
Initial velocity, u = 0
It falls from a height of 3 m and it takes 968 ms to reach the ground.
We need to find the acceleration due to gravity on Gazorpzorp. Using second equation of motion to find a. It can be given by :
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
u = 0 (at rest) and a = g
[tex]s=\dfrac{1}{2}at^2\\\\a=\dfrac{2s}{t^2}\\\\a=\dfrac{2(3)}{(968\times 10^{-3})^2}\\\\a=6.4\ m/s^2[/tex]
So, the acceleration due to gravity on Gazorpzorp [tex]6.4\ m/s^2[/tex].
