What is the magnitude of the brick's velocity just before it reaches the ground?

Question

siyuation: A brick is dropped (zero initial speed) from the roof of a

building. The brick strikes the ground in 2.50 s. You may ignore air resistance, so the brick is in free fall.

Answer 1

v = at

a = 9.8 m/s^2

t = 2.5 s

v = 9.8 m/s^2 (2.5 s)

= 24.5 m/s

that's it!

Answer 2

Your working formula is

Vf - Vo = gT

where

Vf = velocity of brick just before it reaches the ground

Vo = initial velocity = 0 (since brick was dropped)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

T = time of fall = 2.5 sec (given)

Substituting values,

Vf - 0 = 9.8(2.5)

Vf =24.5 m/sec.

Answer 3

distance=Vit+at^2/2

distance=(9.8)(2.5)^2/2

distance=(9.8)(6.25)/2

distance=30.625 m

Vf^2=Vi^2+2ad

Vf^2=2ad

Vf^2=2(9.8)(30.625)

Vf^2=(19.6)(30.625)

Vf^2=600.25

Vf=sqrt.(600.25)

Vf=24.5 m/s

Answer 4

v=u+at

v = 0+9.8*2.5

v = 24.5m/s