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Homework Problems

  1. Oct 16, 2003 #1
    Hi,

    We were assigned a problem set for homework, a total of 7 problems. Below are the questions, and the work I used to get the answer. I'm just looking for a second set of eyes to point out any mistakes I may have made.

    thanks!

    =======
    6.(a) How long will it take a freely falling body to acquire a speed of 58.8 m/s? (b) How far will it have fallen during that time?

    a) t = v/a t = 58.8 / 9.8 = 6s

    b) d=VoT + 1/2 gt^2
    vo = 58.8 m/s
    vf =
    a = 9.8 m/s
    t = 6 secs
    d = 529.2 <=== Answer

    10) A coin dropped from the top of a precipice takes 5.00 s to hit the ground. How high is the precipice?

    d = 1/2 at^2
    d = 1/2 * 9.8 * 5.00
    = 24.5 m

    11) A ball thrown vertically upward returns to the hand of the thrower 6.00 s later. (a). For how many seconds did the ball fall after reaching its high point? (b) How high did the ball go?

    a) t = 2d/a
    t = 2 * 176.4/9.8 = 36 = 6

    b) d = 1/2 at^2
    d = 1/2 * 9.8 * 6.0s^2
    = 176.4 m

    13) An object falls to the floor from a shelf 2.5 m high. With what speed does it strike the floor?

    2.5 m * 9.8
    = 24.5 m/s

    16) (a) With what speed does a freely falling body dropped from a height of 88.2 m hit the ground? (b). How long does the body take to fall this distance?

    a) v^2 = 2ad
    v^2 = 2(9.8)(88.2)
    = 1728.72

    b) t = 2d/a t = 2(88.2)/9.8 = 18s

    18) A bullet shot vertically upward has an initial speed of 588 m/s. (a). How long does it take before the bullet stops rising? (b). How high does the bullet go during this time?

    a) t = v/a t = 588/9.8 = 60s

    b) d = 1/2 at^2
    d = 1/2 (9.8) (60)^2
    = 1760 m

    20) An object dropped from a ballon descending at 4.2 m/s lands on the ground 10 s later. What was the altitude of the balloon at the time the object was dropped?

    vf = Vo + at
    vf = 4.2 m/s + 9.8 * 10s
    = 102.2 m
     
  2. jcsd
  3. Oct 16, 2003 #2

    Hurkyl

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    What is the initial velocity and final velocity?

    Oh, and just to be picky (and because I think correct notation helps prevent mistakes), the formulas you used should be:

    t = &Delta;v / a
    and
    &Delta;d = v0 t + (1/2) a t2

    (where &Delta;v means the change in v, and similarly for d)

    And just to make sure you're aware, did you intend to select downwards as the positive direction in this and later problems? (There's nothing inherently wrong with such a selection, but it is often a symptom of an underlying mistake)


    Well, this can't possibly be right.

    2.5 m * 9.8 m/s2 = 24.5 m2/s2... the units don't work out! You know the right formula (you used it in another problem), so look back at what you know and pick the right formula!


    What is the full formula you used for part (b)? You got the right answer (I think), I just want to make sure it wasn't due to compensating errors.


    You were solving for vf, but the question asks for displacement!!

    A helpful hint is to always handle your units with care... in this case, though you didn't notice you were solving for the wrong variable, if you were careful about units you would have noticed that your calculatiins gave you m/s when you were looking for an m!


    Everything else looks right.
     
  4. Oct 16, 2003 #3
    #6 is straight from the book. It doesn't list, nor ask for initial and final velocity. Am I supposed to find these? Or are my answers correct for what was given?

    As for selecting downward as the positive direction... Can you elaborate a bit more on that? He gave us 7 equations to work with for these and other problem sets. Just using what fits.

    As for 13 would it be:

    V^2 = 2ad
    V^2 = 2(9.8)(2.5)
    = 49

    For 18(b) that is the full formula. The formula is d = 1/2(a)(t)^2

    For 20, would it be:

    d = v0 t + (1/2) gt^2
    d = 4.2 m/s * 10 + (1/2) * 9.8 * 10^2
    = 532 m
     
    Last edited by a moderator: Oct 16, 2003
  5. Oct 17, 2003 #4

    HallsofIvy

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    Yes, you were not asked for that- but it would have indicated more that you knew what you were doing if you had given them. Hurkyl SAID he was being picky! His point was that the formula you used "t= v/a" looks like v is "velocity". Actually, as he said, it should be "change in velocity" (which is why he said it would be &DELTA;v rather than v). You are asked how long it will "acquire a speed of 58.8 m/s". The figure given in the problem IS change in velocity only if you assume the initial velocity is 0. That wasn't actually said but it's a reasonable assumption given that it was not said.

    Did you think about which direction was positive? "Applications" don't come with coordinate systems attached! You have to select one yourself and its a good idea to be aware of exactly what you are doing. Most people tend to choose positive as "upward" and so write the acceleration due to gravity as -9.8 m/s2. As long as you are consistent, you are okay.

    Hurkyl's point was the obvious fact that you can't get velocity by multiplying distance and acceleration! The formula you have now is good. (Do you recognize it as "conservation of energy"? (1/2)mV2 is kinetic energy and mad is potential energy. At the top total energy is potential energy only, at the bottom kinetic energy only so the two must be the same. Divide both sides of the equation by 2m to get your formula.)

    [/quote]For 18(b) that is the full formula. The formula is d = 1/2(a)(t)^2[/quote]

    Do you mean that this formula was given as part of the problem? Surely it should be d= d0+ v0t+ (1/2)at2 where d0 is the initial height and v0 is the initial velocity. If you are measuring from the initial position, as in this problem, then d0= 0. If, in addition, the initial velocity is 0, then v0= 0 and the formula you have is correct. Here that is NOT true. The initial velocity is given as 588 m/s so your height formula should be
    d= -588 t+ (1/2)(9.8) t2. Do you see why I have taken v0 to be negative?? Would you be surprised if d were negative?

    This is the distance the object fell. Is that what the question asked for?
     
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