1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Due Tonight (jan 27)

  1. Jan 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi im just starting a college physics class, all the work is pretty much online and im am having a terrible time learning all the equations and things of that nature. The question is

    You are standing by your living room window when an object falls vertically downward in front of you. You observe that it takes 0.210 seconds for the object to pass the window, which is 1.35 m tall. From what height did the object fall from above the window?

    I know that the object is experiencing 9.8 m/s in freefall but i have no clue where to start as far as equations go. Please help im terrible at this.
    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 27, 2009 #2
    Im sorry that i cant really provide any work, im definately not trying to freeload, and i do want to learn how to do this stuff. I just need someone to walk me through it.
  4. Jan 27, 2009 #3


    User Avatar
    Homework Helper

    Welcome to PF.

    You just need to know that the distance an object falls is given by:

    x = 1/2*g*t2

    Since you know the difference in time and you know the distance it traveled in that additional time then you just subtract one equation for one distance from the other. And since 1 time is longer by the time they give you ... then just solve the resulting quadratic equation.
  5. Jan 27, 2009 #4
    what is the g in the equation? im guessing x would be the height and t is the time
  6. Jan 27, 2009 #5
    im not too sure, what i would probably work towards would be:
    well it took the object .210 seconds to move 1.35 meters, so find out that initial velocity (when it first is visible from the window)

    then you can use the velocity you solved for (when the object is at the top of the window/ first visible) and use that as the final velocity again, and solve for the distance.

    edit: g is gravity, and lowlys method is good too.
  7. Jan 27, 2009 #6
    Already answered.
    Last edited: Jan 27, 2009
  8. Jan 27, 2009 #7
    so my equation should look like this

    x=1/2 * 9.8m/s * .210 Squared
  9. Jan 27, 2009 #8
    V= 9.8m/s Squared *.210 gives me 2.058m/s squared thats using the forumula V=gt is this correct for the initial velocity?
  10. Jan 27, 2009 #9
    still need some help please
  11. Jan 27, 2009 #10


    User Avatar
    Homework Helper

    No. The .210 is the time difference.

    X1 = 1/2*9.8*t12
    X2 = 1/2*9.8*t22

    And you know that t2 = t1 + .210
    And x2 - x1 = 1.35

    The rest is algebra.
  12. Jan 27, 2009 #11
    im still lost sorry i must just be stupid
  13. Jan 27, 2009 #12
    how do i go about finding what t1 is?
  14. Jan 27, 2009 #13


    User Avatar
    Homework Helper

    Sorry, I won't solve it for you.

    It won't do either of us any good.

    Just keep trying.
  15. Jan 27, 2009 #14
    ok then can you explain it in a different way

    i have no idead how to find t1
  16. Jan 27, 2009 #15
    ok heres what im trying to think of in my head. Basically the question states that it takes a object .210seconds to cover 1.35meters. So with that given information i have to find what height the object falls from. Well a given is the objects acceleration is 9.8m/s squared
  17. Jan 27, 2009 #16
    ok i may be stupid or something, this is suppose to be easy and i ve been working on it for like a hour please help
  18. Jan 27, 2009 #17
    ok i cam up with this. Will this equation work?

    D=Vi *t +1/2 * a * t2

  19. Jan 27, 2009 #18
    am i that lost that nobody can help me?
  20. Jan 27, 2009 #19
    i guess it would help if i found the avg speed which is 1.35/.210 = 6.43m/s
  21. Jan 27, 2009 #20
    AVG Speed = 1.35/.210 = 6.43 m/s
    a = (vm - vt)/t
    9.81 = (6.43 - vt)/(0.105)

    vt = 5.40 m/s

    5.402 = 0 2 + 2(9.81)d

    d = 1.49 m
    Object was released 1.49 m above window.

    is this correct?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?