Question

A 200 kg boulder is 1000m above the ground. What is the speed of the object right before it hits the ground?

Answer 1

`140.07m/s`

Step-by-step explanation:

we use the equation for the final speed:

`v_(f)^2=v_(0)^2+2gh`

where
`v_(f)` is the final speed,
`v_(0)` is the initial speed,
`g` is the gravitational acceleration (
`g=9.81m/s^2`) and
`h` is the height.

In this case we don't have an initial velocity indicated so:
`v_(0)=0`, and we are told that the boulder is at a height:
`h=1,000m`.

We substitute this values into the equation:

`v_(f)^2=2gh\\v_(f)^2=2(9.81m/s^2)(1,000m)\\v_(f) ^2=19,620m^2/s^2\\\\v_(f)=√(19,620m^2/s^2) \\v_(f)=140.07m/s`

the speed of the object right before it hits the ground is
`140.07m/s`