Question

An object is dropped at from rest from he top of a 100m building.How long will it take for the object to hit the ground?

Answer 1

Do you know which equations of motion to use?

Answer 2

\[\Delta y=Vt+\frac{ 1 }{ 2 } g t ^{2}\]

Answer 3

Vt = 0 because your initial velocity is 0, g = -9.8 m/s^2, solve for t.

Answer 4

what does that first part say change in what? what does y stand for?

Answer 5

y is your change in height. It will be -100m since your initial height is 100m and your final height is 0m.

Answer 6

i wrote down iv=o
t=?
d=100m
g=9.8m/s^2

Answer 7

I prefer to use y instead of d in this case because it is a vertical change, not a horizontal.

Answer 8

So, you have -100m=(0 m/s)(t)-(1/2)(9.8m/s^2)(t^2)

Answer 9

This simplifies to -100=-4.9t^2

Answer 10

Divide and then take the square root of 100/4.9. It should be close to the square root of 20.

Answer 11

i got 2.04 seconds

Answer 12

That sounds about right. =)

Answer 13

thanks:)i c in this example it has 4.52 seconds

Answer 14

And it followed the same "path" we chose, right?

Answer 15

yea i understand your way but i dnt understand why it has that answer

Answer 16

What does the question say exactly? Sometimes leaving something small out might change the way the question is answered.

Answer 17

it has d=vit+1/2at^2
d=1/2 at^2
t=square root 2d/a
t=square root 2(100)/9.81=4.52 s

Answer 18

I just did it out the way I showed you and I got the same answer as the book. I did not get my calculator out before. They way they do it here is instead of multiplying the 9.8 by 0.5 to get 4.9, they multiplied the 100 by 2 instead and then divided by 9.8. The answer works out the same both ways, but I think there may have been an error in the input process on the calculator.

Answer 19

so you think your answer would still be correct right?

Answer 20

Compared to what the book gave, I got the same 4.52 seconds.

Answer 21

What did you put into the calculator?

Answer 22

D = Vit + 1/2at^2
Starting from rest so Vit can be removed (initial velocity * time)
D = 1/2at^2
100m= 1/2at^2 (a = acceleration = gravity = 9.8 m/s^2)
100m=4.9 m/s^2 * t^2
20.408 s^2 = t^2
t = 4.5175s

Answer 23

Thank you for writing that out greenleaf. =P

Answer 24

No problem

Answer 25

k i did
100/24.01=sqrt 4.16......2.04 @greenleaf800073 for you r help and thanks @Hagopian13

Answer 26

Where did the 24.01 come from?

Answer 27

Ooooooohhhh, I see.

Answer 28

You accidentally squared the 4.9 m/s^2. That does not need to squared as the time t is the only part of the equation being squared. That is why your answer worked out to the 2.04 seconds.

Answer 29

oh ok .u c where it says 100m=4.9m/s^2x t^2....the 20.408 u got from where

Answer 30

The 20.48 is from 100/4.9.

Answer 31

yea n the 4.517 is frrom/?

Answer 32

Taking the square root of 20.48.

Answer 33

K THANKS ALOTi get it

Answer 34

Whoop whoop! =)