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

Apr 11, 2016

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 ${v}^{2} = {u}^{2} + 2 a s$

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:
$F = w = m g$
Where
g is the gravitational field strength. Near the Earth's surface the value of g* is $9.8 N . k {g}^{- 1}$

Newton's second law tells us that resultant force is equal to the product of mass and acceleration:
$F = m a$

Combine the two equations:
$F = m g = m a$
Mass cancels out to leave: $a = g = 9.8 m . {s}^{- 2}$