Results 1 to 4 of 4

Math Help - [SOLVED] Physics problem w/ guns

  1. #1
    Junior Member
    Joined
    Aug 2008
    Posts
    53

    [SOLVED] Physics problem w/ guns

    Two identical pellet guns are fired simultaneously from the edge of a cliff. These guns impact an initial speed of 30.0 m/s to each pellet. Gun A is fired straight upward, with the pellet going up and then falling back down, eventually hitting the ground beneath the cliff. Gun B is fired straight downward.


    In the absence of air resistance, how long after pellet B hits the ground does pellet A hit the ground?

    when i look at this i saw that they only gave us 2 pieces of information
    there's gravity and initial velocity i believe
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by john doe View Post
    Two identical pellet guns are fired simultaneously from the edge of a cliff. These guns impact an initial speed of 30.0 m/s to each pellet. Gun A is fired straight upward, with the pellet going up and then falling back down, eventually hitting the ground beneath the cliff. Gun B is fired straight downward.


    In the absence of air resistance, how long after pellet B hits the ground does pellet A hit the ground?

    when i look at this i saw that they only gave us 2 pieces of information
    there's gravity and initial velocity i believe
    Yes, the height of the cliff from the ground below is not that important in this problem.

    This problem is more on your knowlefge of the parabolic flight of the bullet where air resistance is ignored.

    The bullet fierd straight up will reach its max then it will fall down. At the maximum height, the vertical velocity is zero. Then at the same level where it was fired, the vertical velocity is the same as the initial velocity but in opposite direction.
    Here it will be -30m/s.
    That is the same as the initial velocity of the second bullet that is fired straight down.
    That means the second bullet will be ahead of the first bullet by the amount of time that the first bullet spent from cliff level to max height to cliff level again.

    At the max height, the vertical velocity of the first bullet is zero,
    v = Vo -gt
    0 = 30 -gt
    t = 30/g seconds.

    When the bullet falls to the level it was fired, it takes the same t = 30/g seconds again.
    So it takes the first bullet 2(30/g) seconds to be in the same level where the second bullet was fired.

    Therefore, whatever is the height of the cliff from the ground below, the first bullet will always be 60/g seconds late than the second bullet.

    Hence, if your g = 10 m/sec/sec, then the first bullet will be 60/10 = 6 seconds late in hitting the ground.
    If your g = 9.8 m/sec/sec, it would be 60/9.8 = 6.12245 seconds late.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,548
    Thanks
    540
    Hello, john doe!

    Two identical pellet guns are fired simultaneously from the edge of a cliff.
    These guns have an initial speed of 30.0 m/s.
    Gun A is fired straight upward, with the pellet going up and then falling back down,
    eventually hitting the ground beneath the cliff.
    Gun B is fired straight downward.

    In the absence of air resistance, how long after pellet B hits the ground
    does pellet A hit the ground?
    The formula is: . h \;=\;h_o + v_ot - 4.9t^2 . where: . \begin{Bmatrix} h &=& \text{height of pellet} \\ h_o &=& \text{initial height} \\ v_o &=& \text{initial velocity} \end{Bmatrix}

    . . \begin{array}{cccc}\text{Gun A:} & h &= & h_o + 30t - 4.9t^2 \\ \\[-3mm]<br />
\text{Gun B:} & h &=& h_o -30t - 4.9t^2\end{array}



    When does the pellet from Gun A hit the ground?

    . . 4.9t^2 - 30t - h_o \:=\:0 \quad\Rightarrow\quad t_A \;=\;\frac{30 \pm\sqrt{900 + 19.6h_o}}{9.8}\: seconds


    When does the pellet from Gun B hit the ground?

    . . 4.9t^2 + 30t - h_o \:=\:0 \quad\Rightarrow\quad t_B \;=\;\frac{-30 \pm\sqrt{900 +19.6h_o}}{9.8}\: seconds



    Answer: . t_A - t_B \;=\;\left(\frac{30 \pm\sqrt{900+19.6h_o}}{9.8}\right) - \left(\frac{-30 \pm\sqrt{900 + 19.6h_o}}{9.8} \right)

    . . . . . . . . . . . . = \;\frac{60}{9.8} \;=\;\frac{300}{49} \;\approx\; 6.12\text{ seconds}


    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,671
    Thanks
    299
    Awards
    1
    Quote Originally Posted by Soroban View Post
    Hello, john doe!

    The formula is: . h \;=\;h_o + v_ot - 4.9t^2 . where: . \begin{Bmatrix} h &=& \text{height of pellet} \\ h_o &=& \text{initial height} \\ v_o &=& \text{initial velocity} \end{Bmatrix}

    . . \begin{array}{cccc}\text{Gun A:} & h &= & h_o + 30t - 4.9t^2 \\ \\[-3mm]<br />
\text{Gun B:} & h &=& h_o -30t - 4.9t^2\end{array}



    When does the pellet from Gun A hit the ground?

    . . 4.9t^2 - 30t - h_o \:=\:0 \quad\Rightarrow\quad t_A \;=\;\frac{30 \pm\sqrt{900 + 19.6h_o}}{9.8}\: seconds


    When does the pellet from Gun B hit the ground?

    . . 4.9t^2 + 30t - h_o \:=\:0 \quad\Rightarrow\quad t_B \;=\;\frac{-30 \pm\sqrt{900 +19.6h_o}}{9.8}\: seconds



    Answer: . t_A - t_B \;=\;\left(\frac{30 \pm\sqrt{900+19.6h_o}}{9.8}\right) - \left(\frac{-30 \pm\sqrt{900 + 19.6h_o}}{9.8} \right)

    . . . . . . . . . . . . = \;\frac{60}{9.8} \;=\;\frac{300}{49} \;\approx\; 6.12\text{ seconds}


    Nicely done, though I'm going to get picky about one thing: Since 30 < \sqrt{900 + 19.6h_0} for any h_0 > 0 the negative sign before the square root in the t_A - t_B equation is invalid. So it should read
    t_A - t_B \;=\;\left(\frac{30 + \sqrt{900+19.6h_o}}{9.8}\right) - \left(\frac{-30 + \sqrt{900 + 19.6h_o}}{9.8} \right)
    which, of course, leads to the same result Soroban already gave, so this is only a small matter. I include it only for the sake of completeness.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. physics solved with calclus
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 15th 2009, 10:29 PM
  2. [SOLVED] Physics Problem
    Posted in the Advanced Applied Math Forum
    Replies: 6
    Last Post: September 11th 2008, 04:53 AM
  3. [SOLVED] Physics problem
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: September 3rd 2008, 11:39 PM
  4. [SOLVED] Physics problem
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: August 17th 2008, 07:54 PM
  5. physics, acceleration, physics problem
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: September 29th 2007, 03:50 AM

Search Tags


/mathhelpforum @mathhelpforum