Question

A 10 kg rock is suspended 320 m above the ground. Calculate the approximate speed with which the rock hits the ground.

v =

A. 6300 m/s
B. 79 m/s
C. 56 m/s
D. 40. m/s

Answer 1

The answer for this is actually b. 79 m/s

I submitted it to odyssey-ware and i got the question correct

Answer 2

B. 79 m/s

Step-by-step explanation:

Since we have information about the distance of the fall, we can use the following formula to find the final velocity of the object:

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

clearing for
`v_(f)`

`v_(f)=\sqrt{v_(i)^2+2gh}`

where
`v_ {f}` is the final velocity (when it hits the ground),
`v_ {i}` the initial velocity,
`g` is the acceleration of gravity (
`g=9.8m/s^2`), and
`h` is the height from where the rock is dropped, in this case:
`h=320m`

Assuming that the rock was only dropped without adding an initial velocity,
`v_ {i}` is zero.

Substituting all the values in the formula:

`v_(f)=√((0)^2+2(9.8m/s^2)(320m))`

`v_(f)=√(6272)`

`v_(f)=79.2m/s`

The value that best approximates from the options is 79 m/s, the answer is B.