# Homework Help: Last one i have problems with

1. Sep 13, 2011

### Brock Skywalk

1. AN OBJECT DROPPED FROM A WINDOW OF A TALL BUILDING HITS THE GROUND IN 12.0s. IF ITS ACCELERATION IS 9.80m/s^2, THE HEIGHT OF THE WINDOW ABOVE THE GROUND IS?

2.Not sure again

3. Not sure

2. Sep 13, 2011

### lewando

Hello Brock, welcome to PF! Please read and follow the rules, and you will get better attention. Since you are new... There are 4 standard kinematic equations for constant acceleration that you should memorize. You won't memorize them overnight unless you brother calls you Rainman for some reason. Just do as many of these kinds of problems as you can and you will get there someday. They are these (shown for x-direction motion, but same for y-direction with different subscript):

vx = vx0 + axt
x = x0 + (1/2)(vx0 + vx)t
x = x0 + vx0t + (1/2)axt2
vx2 = vx02 + 2ax(x - x0)

So if you know t, ay, vy0, y, and you are looking for y0, which equation shows promise?

3. Sep 14, 2011

### Brock Skywalk

Ok, i am still confused. what do i do first? thanks again for the reply sir. i appreciate it.

-Skywalker

4. Sep 14, 2011

### quietrain

you need to know what the 4 equations stand for first.

v = u + at
v2 = u2 + 2as
s =(1/2) (u+v)t
s = ut + (1/2) at2

to understand these equations you can imagine the following

if skywalker walks a distance "s" in time "t"
his speed at the start of the walk is "u"
and his speed at the end of the walk is "v"
and these 4 equations give the relationship between "s","t","u","v"

so assuming a question gives you "v","u","t" ,"a" but it didn't give you "s",
so you would use the first equation v = u + at, because this equation doesn't involve "s".
and likewise.

5. Sep 14, 2011

### Brock Skywalk

But i am trying to calculate the distance. what is given is (t) which = 12s and a = 12m/s^2. i am not sure which one to apply.

6. Sep 14, 2011

### Brock Skywalk

ok. i got 117.6 M. i used equation #1. what do y'all think?

7. Sep 14, 2011

### lewando

You calculated v, not s.

You know this:
t = 12s (given)
a = 9.81m/s2 (given, though it is really -9.81m/s2 because it is in the downward direction).
u = 0 (given-- the object is dropped, not thrown)

s is what you are looking for--the distance.

So, which of quietrain's excellently simplified and well-explained equations has t, a, and u on the right hand side?

8. Sep 14, 2011

### Brock Skywalk

705.6 M. i used the last one.

9. Sep 14, 2011

### lewando

There you go! Some other things to always do:
Draw a picture of what is going on.
Indicate your coordinate system and location of (x,y) = (0,0).
Use proper signs in front of v, a, even s.

10. Sep 14, 2011

### Brock Skywalk

Thanks Lewando! i really appreciate this. this whole physics thing is a totally different monster. the math is simple is just know how to apply the formulas that i am having trouble with.

so could u explain what the variables stand for? a = acceleration t= time but what do the others really stand for?

11. Sep 14, 2011

### Brock Skywalk

ok. just did another.

an object is thrown vertically upward @ 35m/s. taking g = 10m/s^2, the velocity 5 seconds later is?

i got 85m/s.

i used v = v + a

whatta ya think?

12. Sep 14, 2011

### lewando

I think "v = v + a" is not one of the equations. Precision is the key for good outcomes.

Looks like you used "v = u + at" without regard to the direction of a.

What direction is a acting in?

13. Sep 14, 2011

### Brock Skywalk

yes it was v = v + a*t. what did i do wrong sir?

"a" should be downward, so should i have multiplied by (-10m) thus my answer being 15m/s?

14. Sep 14, 2011

### lewando

Yes!

(v = u + a*t, u being the initial velocity)

15. Sep 14, 2011

### Brock Skywalk

Great! so 15m/s downward?

16. Sep 14, 2011

### lewando

The answer that you should have got was -15m/s [-15 = 35 + (-10)*5]. The negative sign indicates downwardness. I missed that on post 14. You can say 15m/s (a magnitude) downward (a direction) and be correct.

What will get you in trouble is if you say -15m/s downward... it's a double negative of sorts.

17. Sep 14, 2011

### Brock Skywalk

yes sir. i understand now. thank u so much sir. i hope your a sir, lol! i really appreciate it. -Skywalker