Results 1 to 3 of 3

Math Help - Projected particle

  1. #1
    Member
    Joined
    Nov 2009
    Posts
    94

    Projected particle

    Hi
    I have been stumped by this question for the past few days. Worrying since I will be sitting STEP in June! Never mind. It goes like this:
    A particle is projected vertically upwards with a speed of 30m/s from a point A. The point B is h metres above A. The particle moves freely under gravity and is above B for a time 2.4s. Calculate the value of h.
    These are my parameters:
    From A to B
    u=30
    a=-9.8
    s=h
    From B
    u=?
    a=-9.8
    s=?
    t=2.4
    I took t=2.4 as the Maximum height above B.(I am not sure about this line of reasoning!)
    Based on that V=0
    Using V=U + at
    U=23.52 m/s (This seemed to make sense since particle will start decelerating at some point)
    So, U (at b)=23.52 m/s
    Considering it from A to B
    U=30
    V=23.52
    a=-9.8
    S=h
    Using V^2=U^s +2aS
    h=17.7m (3s.f)
    But alas, it seems my answer (and probable reasoning) is wrong. Any suggestions???
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7
    Quote Originally Posted by gbenguse78 View Post
    Hi
    I have been stumped by this question for the past few days. Worrying since I will be sitting STEP in June! Never mind. It goes like this:
    A particle is projected vertically upwards with a speed of 30m/s from a point A. The point B is h metres above A. The particle moves freely under gravity and is above B for a time 2.4s. Calculate the value of h.
    These are my parameters:
    From A to B
    u=30
    a=-9.8
    s=h
    From B
    u=?
    a=-9.8
    s=?
    t=2.4
    I took t=2.4 as the Maximum height above B.(I am not sure about this line of reasoning!)
    Based on that V=0
    Using V=U + at
    U=23.52 m/s (This seemed to make sense since particle will start decelerating at some point)
    So, U (at b)=23.52 m/s
    Considering it from A to B
    U=30
    V=23.52
    a=-9.8
    S=h
    Using V^2=U^s +2aS
    h=17.7m (3s.f)
    But alas, it seems my answer (and probable reasoning) is wrong. Any suggestions???
    The particle moves up from B and returns to B. So the displacement is 0.

    s=v_Bt-\frac{1}{2}gt^2

    0=v_B(2.4)-\frac{1}{2}(9.8)(2.4)^2

    v_B=11.76\ m/s



    v_B^2=v_A^2-2gh

    11.76^2=30^2-2(9.8)h

    h=38.8624\ m
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2009
    Posts
    94
    Quote Originally Posted by alexmahone View Post
    The particle moves up from B and returns to B. So the displacement is 0.

    s=v_Bt-\frac{1}{2}gt^2

    0=v_B(2.4)-\frac{1}{2}(9.8)(2.4)^2

    v_B=11.76\ m/s



    v_B^2=v_A^2-2gh

    11.76^2=30^2-2(9.8)h

    h=38.8624\ m
    Thanks. Much appreciated!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with projected averages
    Posted in the Statistics Forum
    Replies: 0
    Last Post: May 17th 2010, 05:36 AM
  2. Calculating Projected Earnings
    Posted in the Business Math Forum
    Replies: 2
    Last Post: April 9th 2010, 10:40 AM
  3. Replies: 2
    Last Post: April 30th 2009, 04:46 AM
  4. Rotated projected plans
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 16th 2008, 12:58 PM
  5. Particle
    Posted in the Calculus Forum
    Replies: 10
    Last Post: July 11th 2006, 05:01 AM

Search Tags


/mathhelpforum @mathhelpforum