Question

If an object is dropped, how fast will it be moving after falling `3 m`?

Answer 1

The velocity after falling 3m would be `58.8 m.s^(-1)`

If drag is ignored.

Explanation:

All objects in free fall near the Earth's surface and with no drag have a constant acceleration of `9.8 m.s^(-2)` *. To answer the question we will ignore the effects of drag acting on the object (as no information about the drag is provided).

State data and select equation:
`s = 3.0 m`
`u = 0` (anything dropped starts with zero velocity)
`v = ?`
`a = 9.8 m.s^(-2)`
`t = ?`

Use `v^2 = u^2 + 2as`

Substitute values into the equation:
`v^2 = u^2 + 2as = 0 + 2 × 9.8 × 3 = 58.8 m.s^(-1)`

---

The reason for an acceleration of `9.8 m.s^(-2)` is because with no drag the resultant force is equal to the weight only. We can write the following equation:
`F = w = mg` ①
Where g is the gravitational field strength. Near the Earth's surface the value of g* is `9.8 N.kg^(-1)`

Newton's second law tells us that resultant force is equal to the product of mass and acceleration:
`F = ma` ②

Combine the two equations:
`F = mg = ma`
Mass cancels out to leave: `a = g = 9.8 m.s^(-2)`