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

2 Answers
Jun 29, 2018

Answer:

#"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

Answer:

#"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.)