1. Limited time only! Sign up for a free 30min personal 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 Help: Impulse, Speed, and Height

  1. Dec 2, 2009 #1
    Good day, I have a homework problem I'd like help with. I got into getting the impulse (not sure if correct, though), but don't really know how to get the rest of the things. Any tips on what direction I should be going are appreciated.

    1. The problem statement, all variables and given/known data
    Starting from rest, 65kg athlete jumps down onto a platform that is .600m high. While the athlete is on contact with the platform during the time interval [tex]0 < t < 0.800 s[/tex], the force exerted is discribed by the function [tex]F = (9200 N/s)t - (11500 N/s^2)t^2[/tex].

    [tex]m = 65 kg[/tex]
    [tex]h_p = .600 m[/tex](platform)
    [tex]V_0 = ?[/tex] (Is this supposed to be 0? She starts from rest but jumps up and then goes down to the platform)
    [tex]V_f = ?[/tex]
    [tex]h_m = ?[/tex](max)

    (a)What impulse did the athlete receive from the platform?

    (b)With what speed did she reach the platform?

    (c)With what speed did she leave it?

    (d)To what height did she jump upon leaving the platform?

    2. Relevant equations
    [tex]F = (9200 N/s)t - (11500 N/s^2)t^2[/tex]

    [tex]\vec{I} \equiv \int_t^t \Sigma \vec{F}dt[/tex](top t is final, bottom initial...)

    3. The attempt at a solution

    (a)What impulse did the athlete receive from the platform?

    I punched in [tex]F = (9200 N/s)t - (11500 N/s^2)t^2[/tex] values into [tex]\vec{I} \equiv \int_t^t \Sigma \vec{F}dt[/tex]

    [tex]\vec{I} \equiv \int_0^.8 (9200 N/s)(t) - (11500 N/s^2)(t^2) [/tex]

    [tex]\equiv \int (4600 N) (.8 s)^2 - (3833.33 N) (.8 s)^3 [/tex]

    [tex]\equiv (2944 N/s) - (1962.5 N/s)[/tex]

    [tex]\vec{I} \equiv 981.5 N/s [/tex]
  2. jcsd
  3. Dec 2, 2009 #2


    User Avatar
    Homework Helper

    Your work for (a) looks good.
    The question isn't entirely clear, but it must mean she comes to a stop at the end of that 0.8 s interval. If so, you can use impulse = m*Δv to do (b).
  4. Dec 2, 2009 #3
    Using [tex]I = m*\Delta v[/tex]

    I get [tex]\Delta V = 15.1 m/s[/tex]

    [tex]\frac {I}{m} = \Delta v[/tex]

    [tex]\frac {981.5 N/s}{65 kg} \equiv \Delta v[/tex]

    [tex]\Delta v \equiv 15.1 m/s[/tex]

    So Delta V would be the landing speed since when she starts falling down the velocity would be 0. Or would [tex]\Delta v = 15.1 m/s[/tex] be the difference in velocity of when she lands and when she jumps(which means it would count the time she ascends as well)?

    [tex]V_f^2 = V_i^2 +2a(Y_f-Y_i)[/tex]
    vf ^2= vi^2 +2a (yf-yi)
    15.1^2 = 0 + 19.62 (yi)

    Falling from a distance of 11m is kinda impossible isn't it? So 15.1m/s is the difference from when she jump... right?
    Last edited: Dec 2, 2009
  5. Dec 2, 2009 #4


    User Avatar
    Homework Helper

    There is no mention of a jump in the question.
    If she does jump, how do we know what part of the impulse stops the fall and what part makes the jump? Maybe you should write out the whole question - perhaps we are missing some more information.
  6. Dec 2, 2009 #5
    Is what's on the book. I too, find it not to make much sense. :(
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook