Question

On planet Gazorpzorp, an object from rest at a height of 3.00m takes 968 ms to reach the ground. What is the acceleration due to gravity on Gazorpzorp. Please show your work if possible. ​

Answer 1

`a=6.4\ m/s^2`

Step-by-step 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 :

`s=ut+(1)/(2)at^2`

u = 0 (at rest) and a = g

`s=(1)/(2)at^2\\\\a=(2s)/(t^2)\\\\a=(2(3))/((968* 10^(-3))^2)\\\\a=6.4\ m/s^2`

So, the acceleration due to gravity on Gazorpzorp
`6.4\ m/s^2`.