What is the Height of a Window Above the Ground?

[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

your assistance is appreciated. -Skywalker

 

[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):

v_x = v_x0 + a_xt
x = x₀ + (1/2)(v_x0 + v_x)t
x = x₀ + v_x0t + (1/2)a_xt²
v_x² = v_x0² + 2_ax(x - x₀)

So if you know t, a_y, v_y0, y, and you are looking for y₀, which equation shows promise?

 

[Brock Skywalk]

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

-Skywalker

 

[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.

 

[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.

 

[Brock Skywalk]

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

 

[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?

 

[Brock Skywalk]

705.6 M. i used the last one.

 

[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.
Check if your answer makes any sense, if possible. Like units.

 

[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?

 

[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 you think?

 

[lewando]

.

i got 85m/s.

i used v = v + a

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?

 

[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?

 

[lewando]

Yes!

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

 

[Brock Skywalk]

Great! so 15m/s downward?

 

[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.

 

[Brock Skywalk]

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