Results 1 to 6 of 6

Thread: Physics: Motion problem

  1. #1
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Physics: Motion problem

    No Calculus allowed. I need some help. I'm really stuck...

    Thank You.


    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    no calculus, huh?

    since $\displaystyle \Delta y = 0$

    $\displaystyle 0 = v_{yo}t - \frac{1}{2}gt^2$

    $\displaystyle 0 = t\left(v_{yo} - \frac{1}{2}gt\right)$

    $\displaystyle v_{yo} = \frac{1}{2}gt$

    $\displaystyle t = \frac{2v_{yo}}{g} = \frac{2v_o\sin{\theta}}{g}$

    $\displaystyle \Delta x = v_x \cdot t$

    $\displaystyle \Delta x = v_o\cos{\theta} \cdot \frac{2v_o\sin{\theta}}{g}$

    $\displaystyle \Delta x = \frac{v_o^2 \cdot 2\sin{\theta}\cos{\theta}}{g}$

    $\displaystyle \Delta x = \frac{v_o^2 \sin(2\theta)}{g}$

    max value for $\displaystyle \Delta x$ occurs when $\displaystyle \sin(2\theta) = 1$

    $\displaystyle \Delta x_{max} = \frac{v_o^2}{g}$


    min ball speed will occur when it is at the peak of its trajectory ... when $\displaystyle v_y = 0$ ... min speed would therefore be ... ?


    max height occurs at $\displaystyle t = \frac{v_{yo}}{g}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1
    Quote Originally Posted by skeeter View Post
    no calculus, huh?

    since $\displaystyle \Delta y = 0$

    $\displaystyle 0 = v_{yo}t - \frac{1}{2}gt^2$

    $\displaystyle 0 = t\left(v_{yo} - \frac{1}{2}gt\right)$

    $\displaystyle v_{yo} = \frac{1}{2}gt$

    $\displaystyle t = \frac{2v_{yo}}{g} = \frac{2v_o\sin{\theta}}{g}$

    $\displaystyle \Delta x = v_x \cdot t$

    $\displaystyle \Delta x = v_o\cos{\theta} \cdot \frac{2v_o\sin{\theta}}{g}$

    $\displaystyle \Delta x = \frac{v_o^2 \cdot 2\sin{\theta}\cos{\theta}}{g}$

    $\displaystyle \Delta x = \frac{v_o^2 \sin(2\theta)}{g}$

    max value for $\displaystyle \Delta x$ occurs when $\displaystyle \sin(2\theta) = 1$

    $\displaystyle \Delta x_{max} = \frac{v_o^2}{g}$


    min ball speed will occur when it is at the peak of its trajectory ... when $\displaystyle v_y = 0$ ... min speed would therefore be ... ?


    max height occurs at $\displaystyle t = \frac{v_{yo}}{g}$

    How did you figure all of these side details out like delta y = 0?


    How do you determine which parts of the problem belong to X and which ones belong to y ?

    For the last part can you at least show me the equations that I would use?

    Thanks,
    qbkr21
    Last edited by qbkr21; Jan 31st 2009 at 02:15 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    How did you figure all of these side details out like delta y = 0?

    if the projectile starts on the ground and ends on the ground, what is $\displaystyle \Delta y$?


    How do you determine which parts of the problem belong to X and which ones belong to y ?

    I don't understand the question ... what parts do you mean?

    For the last part can you at least show me the equations that I would use?

    I gave you the equation for time to the top of the projectile's trajectory ... you need to calculate $\displaystyle \Delta y$ for that time.

    $\displaystyle \Delta y = v_{yo}t - \frac{1}{2}gt^2$


    you could also ignore the time element and use this equation to calculate $\displaystyle \Delta y$ ...

    $\displaystyle v_{yf}^2 = v_{yo}^2 - 2g(\Delta y)$

    at the top of its trajectory, $\displaystyle v_{yf} = 0$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    RE:

    Question:

    b) What is the minimum speed of the ball during this hole-in-one shot?

    The book says the answer is 21.2 m/s

    I set up my equation as follows...

    (V_y)^2=(V_xy)^2-2g*delta(y)
    0=V_xy)^2-1799.28
    x^2=1799.28
    x=42.4

    ***Where did I go wrong?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    at the top of its trajectory, the projectile's y-component of velocity is zero ... so, it only has an x-component of velocity which is constant since there is no acceleration in the x-direction.

    $\displaystyle v_x = v_o\cos{\theta}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Physics motion problem using two dimensions
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: Mar 8th 2010, 08:08 AM
  2. Physics motion problem
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: Jan 24th 2009, 05:57 AM
  3. physics motion
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 27th 2008, 09:57 PM
  4. physics motion problem
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: Jul 12th 2007, 05:29 PM
  5. physics motion
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Jun 24th 2007, 10:07 AM

Search Tags


/mathhelpforum @mathhelpforum