Results 1 to 8 of 8

Math Help - kinematics question

  1. #1
    Senior Member
    Joined
    Jul 2006
    Posts
    364

    kinematics question

    A balloon is rising vertically at a constant speed of 21 m/s. A stone is dropped from the balloon when it is h metres above the ground. The stone strikes the ground 10 seconds later. Assuming that air resistance is negligeble, what is h?

    this probably isnt a very complex question, im probably just forgetting something trivial...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by scorpion007 View Post
    A balloon is rising vertically at a constant speed of 21 m/s. A stone is dropped from the balloon when it is h metres above the ground. The stone strikes the ground 10 seconds later. Assuming that air resistance is negligeble, what is h?

    this probably isnt a very complex question, im probably just forgetting something trivial...
    note:upwards is positive

    (starting height) = (ending height) - (starting velocity)(time) + (1/2)(acceleration)(time)^2

    thus: h = 0 - (21)(10) - (1/2)(-9.8)(10)^2

    then: h = (1/2)(9.8)(100) - 210

    multiply: h = 490 - 210

    finally: h = 280

    Although in your work you should label each number (with meters, seconds etc..)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2006
    Posts
    364
    thank you very much, i wasnt aware of this forumula, as my math text book doesn't cover it. It appears to only have the 5 classic ones.
    EDIT:
    although, why did you go from
    (starting velocity)(time) + (1/2)(acceleration)(time)^2
    to:
    (21)(10) - (1/2)(-9.8)(10)^2

    i.e. why did you change the + to a - ?

    Also, could i have solved it using the classic formula's? such as: s = ut + 1/2at^2 ?
    Last edited by scorpion007; October 14th 2006 at 12:01 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by scorpion007 View Post
    thank you very much, i wasnt aware of this forumula, as my math text book doesn't cover it. It appears to only have the 5 classic ones.
    EDIT:
    although, why did you go from
    (starting velocity)(time) + (1/2)(acceleration)(time)^2
    to:
    (21)(10) - (1/2)(-9.8)(10)^2

    i.e. why did you change the + to a - ?
    I defined upwards to be positive, the rock started out going 21 m/s up, thus it was positive

    However, gravity is going downward, thus the 9.8 is negative.

    Also, could i have solved it using the classic formula's? such as: s = ut + 1/2at^2 ?
    probably, but you'll have to tell me what s and u stand for...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,846
    Thanks
    321
    Awards
    1
    Quote Originally Posted by scorpion007 View Post
    A balloon is rising vertically at a constant speed of 21 m/s. A stone is dropped from the balloon when it is h metres above the ground. The stone strikes the ground 10 seconds later. Assuming that air resistance is negligeble, what is h?

    this probably isnt a very complex question, im probably just forgetting something trivial...
    What Quick did was okay, but he got some postives and negatives screwed up. I'm not razzing him for it, it's a common mistake.

    For starters, Quick WAS using your formula. Apparently he didn't know the variables so he used words instead.
    (Edit: I looked at Quick's work again and, for some reason, the equation he was using has an extra negative sign all throughout it. Mathematically it is the same as the displacement equation used here, but in terms of the Physics I can't make any sense out of it.)

    Let me run you through this the way I would set it up.

    Picture: We have a balloon rising at a velocity of 21 m/s upward. A rock is dropped from rest with respect to the balloon, so the initial velocity of the rock is the same as the balloon's: 21 m/s upward. The rock hits the ground at time t = 10 s.

    Coordinate system: I'm going to choose +y upward and I'm going to choose the origin (y = 0 m) as being on the ground.

    I'm going to use my variation on s = ut + (1/2)at^2. s is a displacement and as we are looking for distances I'm going to write s = y - y0, where y(t) is the height of the rock at a given time. Thus y = y(t) and y0 = y(t = 0). u represents the initial velocity of the object. Usually that is written as u = v0. So the equation becomes:
    y - y0 = v0*t + (1/2)at^2
    I don't know how your textbook presents this equation (presumably s = ut + (1/2)at^2) but this form of the equation is rather more common.

    So. We know:
    t = 10 s
    y = 0 m (the rock is on the ground at t = 10 s)
    y0 = ? (This is what we are looking for)
    v0 = 21 m/s
    a = -g = -9.8 m/s^2 (g is NEVER negative!)

    0 - y0 = 21*10 + (1/2)(-9.8)(10)^2

    -y0 = -280

    y0 = 280 m.

    -Dan
    Last edited by topsquark; October 14th 2006 at 07:41 AM. Reason: Addendum
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by topsquark View Post
    What Quick did was okay, but he got some postives and negatives screwed up. I'm not razzing him for it, it's a common mistake.

    For starters, Quick WAS using your formula. Apparently he didn't know the variables so he used words instead.
    (Edit: I looked at Quick's work again and, for some reason, the equation he was using has an extra negative sign all throughout it. Mathematically it is the same as the displacement equation used here, but in terms of the Physics I can't make any sense out of it.)
    Quote Originally Posted by topsquark View Post
    If we assume that the acceleration is constant the equation of motions are:
    y = y_0 + v_0t + (1/2)at^2
    Quote Originally Posted by topsquark View Post
    y and y0 are the final and initial positions (y = y(t), y0 = y(0).)
    I was solving for y_0,

    so -y_0 = v_0t + (1/2)at^2 - y

    multiply by -1 to get: y_0 = y - v_0t - (1/2)at^2
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,846
    Thanks
    321
    Awards
    1
    Quote Originally Posted by Quick View Post
    I was solving for y_0,

    so -y_0 = v_0t + (1/2)at^2 - y

    multiply by -1 to get: y_0 = y - v_0t - (1/2)at^2
    Ah, I thought that was your starting equation. That makes sense then.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Jul 2006
    Posts
    364
    i see, thanks for all the help guys
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Kinematics question.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 6th 2011, 02:58 PM
  2. Need help with a kinematics question
    Posted in the Calculus Forum
    Replies: 5
    Last Post: March 17th 2011, 12:15 AM
  3. Kinematics Question
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: April 18th 2010, 08:25 AM
  4. Kinematics question
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: October 8th 2009, 04:34 PM
  5. Kinematics question
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: July 12th 2009, 11:09 PM

Search Tags


/mathhelpforum @mathhelpforum