Question

A brick is dropped from a 5 m tall building. At 0.50 s how fast is the brick moving?

Answer 1

The brick is dropped so it's in free-fall meaning \(\large v_i=0\space\space m/s\).
If you apply \(\large v_f=v_i+at\), you'll get your answer.

Answer 2

would it be 5

Answer 3

Nope

Answer 4

k i know the initial v is 0 and vf= 0+1/2 X 9.8?

Answer 5

?
Where did 1/2 come from?

Answer 6

i have a formula that has 1/2 xgravity

Answer 7

OH Sorry. I thought t=0.0 s

Answer 8

My bad

Answer 9

no prob :)

Answer 10

Yes if you round 4.9 then it's 5m.
Just a word of advice. Conventionally, acceleration is written first, not time.

Answer 11

0.50+1/2(9.8)?

Answer 12

No, initial velocity is still 0.
\(v_f=at=gt\)

Answer 13

i have iv =0 vf=0.50

Answer 14

its 10m/s

Answer 15

???
Umm
Initial velocity=0m/s
Acceleration=g=9.8m/s^2
t=0.5s
\(v_f = 0m/s+(9.8m/s^2)(0.5s)=4.9m/s\)

Answer 16

ohhhh thanks i was looking at the wrong formula smt.so basically for this we are trying to the the final velosity

Answer 17

if i have a question like this can i apply the same formula.... how faw will a brick starting at rest fall freely in 5.0 s?

Answer 18

If you want to know "how far", then you're looking for position, not velocity which is totally different

Answer 19

distance deals with how far an object has moved so i can use that instead?

Answer 20

that's the same thing i just said....

Answer 21

k so distance is speed x time so 5.0x 0

Answer 22

\(\huge x_f=x_i+v_it+{1\over 2}at^2\)

Answer 23

its 49 m