Question

If an object is dropped, how fast will it be moving after falling `21 m`?

Answer 1

`"20.3 m/s"`

Explanation:

Use equation of motion

`"v"^2 - "u"^2 = "2aS"`

Where

• `"v ="` Final velocity
• `"u ="` Initial velocity
• `"a ="` Acceleration
• `"S ="` Displacement

The object is dropped. So `"u"` is zero.

`"v" = sqrt(2"aS")`

`color(white)("v") = sqrt(2 × "9.8 m/s"^2 × "21 m")`

`color(white)("v") = 20.3\ "m/s"`

Answer 2

`"20.3 m s"^-1` (to 3 significant figures)

Explanation:

• Using the formula:

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

where;

• `v` is the final velocity of the object which we need to find.
• `u` is the initial velocity of the object, which is `0` since it fell from rest.
• `a` is the acceleration of free fall `("9.81 m s"^-2)`
• `s` is the displacement, which is `"21 m"`

Now all you gotta do is substitute:

`v^2= 0^2+ (2xx"9.81 m s"^(-2)xx"21 m")`

`v^2 = "412.02 m"^2"s"^(-2)`

So

`v= "20.3 m s"^-1` (to 3 s.f.)