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

2 Answers
Jun 29, 2018

"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"

Jun 29, 2018

"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.)