Question

A body of mass 2kg is dropped from rest position 5m above the ground. what is its velocity at height 3.0m above the ground approximately?

Answer 1

`vecv_1 = 6m/s`

Explanation:

Givens:
`vecv_0 = 0m/s`
`vecv_1` = ?
`veca` = 9.8 `m/s^2` [D]
`Deltavecd` = 3m-5m = -2m

Using one of the five equations of motion, we plug in our numbers and solve.

`vecv_1^2 = vecv_0^2+2vecaDeltavecd`

Since our intial velocity is `0m/s`:
`vecv_1^2 = cancelvecv_0^2+2vecaDeltavecd`
`vecv_1^2 = 2(-9.8m/s^2)(-2m)`
`vecv_1^2 = 39.2m^2/s^2`
`sqrtvecv_1^2 = sqrt39.2m^2/s^2`
`vecv_1 = 6.2609...m/s`

With the given significant figures, the final answer is:
`vecv_1 = 6m/s`