Results 1 to 9 of 9

Math Help - Physics problem, hairy algebra

  1. #1
    Junior Member
    Joined
    Sep 2010
    From
    Oslo
    Posts
    57
    Thanks
    4

    Physics problem, hairy algebra

    We have a mini catapult that fires balls at a velocity of 5 m/s. It is placed 1 meter above the ground and pointed at an angle \theta towards the positive x-axis. What angle(s) do we need to set it to in order to hit a target on the ground at x=2.7 meters?

    In other words:

    v_{0}=5 m/s
    x(t)=2.7 m
    y(t)=-1 m (assuming the catapult is standing in the origin)
    \theta=?

    My thinking so far


    First express the velocity in the x-direction and y-direction separately:

    v_{x}=v_{0}cos\theta

    v_{y}=v_{0}sin\theta-gt

    Then find a way to express t in terms of the angle and the positions:

    x(t)=\frac{v_{x}}{t}

    t=\frac{v_{x}}{x(t)}=\frac{v_{0}cos\theta}{x(t)}

    Use what I now have to write y as a function of theta:

    y(t)=v_{y}-\frac{1}{2}gt^2

    y(t)=v_{0}sin\theta-gt-\frac{1}{2}gt^2

    y(\theta)=v_{0}sin\theta-g(\frac{v_{0}cos\theta}{x(t)})-\frac{1}{2}g(\frac{v_{0}cos\theta}{x(t)})^2

    Since I know x, y and v, I should now be able to solve for theta.

    Have I set this up correctly? Is there maybe an easier way to calculate the angle? And finally, if this is the best way to find it, how the hell do I solve for theta??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    Um... the equation for the v_y gives the actual vertical velocity at time t, not the vertical component of the initial velocity.

    And then, you'd also like to get the displacement of the ball in terms of vertical and horizontal components.

    Along the horizontal, you have:

    s_x = v_o\cos\theta t

    Vertically;

    s_y = s_o + v_o \sin\theta t - \dfrac12 gt^2

    Horizontally, s_x = 2.7 m

    Vertically, s_y = -1 m

    You should be able to get the angle more easily like this.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2010
    From
    Oslo
    Posts
    57
    Thanks
    4
    So I'm supposed to get one angle from each equation by substituting in my expression for t?

    t=\frac{v_{x}}{s_{x}}=\frac{v_{o}cos\theta}{s_{x}}

    I did that, and got the 57 degrees from the first one, but the second one I still can't solve..
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    Um...

    The two equations that you get are:

    -1 = 0 + 5 \sin\theta t - \dfrac12 gt^2

    2.7 = 5 \cos\theta t

    From that you get:

    t = \dfrac{2.7}{v_x} = \dfrac{2.7}{5\cos\theta}

    Hm... this will be a little more difficult to solve than I expected, since you get a quadratic. I forgot the -1 would be in the quadratic.

    EDIT:

    Also remember that Speed = Distance/Time.

    So, this : t = v/s is false, but t = s/v is true.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member BAdhi's Avatar
    Joined
    Oct 2010
    From
    Gampaha, Sri Lanka
    Posts
    252
    Thanks
    6
    -1 = 0 + 5 \sin\theta t - \dfrac12 gt^2 .................(1)

    t = \dfrac{2.7}{v_x} = \dfrac{2.7}{5\cos\theta} ..................(2)

    substitute t on the (1) from (2)

    you'll find a quadric equation of the variable \tan\theta
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2010
    From
    Oslo
    Posts
    57
    Thanks
    4
    Great! Thanks guys. Finally I know what equation I need to solve. I'll give it a shot right away.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,108
    Thanks
    970
    \Delta x = v_0 \cos{\theta} \cdot t

    t = \dfrac{\Delta x}{v_0 \cos{\theta}}

    \Delta y = v_0 \sin{\theta} \cdot t - \frac{1}{2} gt^2

    \Delta y = v_0 \sin{\theta} \cdot \dfrac{\Delta x}{v_0 \cos{\theta}} - \frac{1}{2} g \left(\dfrac{\Delta x}{v_0 \cos{\theta}}\right)^2

    \Delta y = \Delta x \tan{\theta} - \dfrac{g \Delta x^2}{2v_0^2} \sec^2{\theta}

    \Delta y = \Delta x \tan{\theta} - \dfrac{g \Delta x^2}{2v_0^2} (1 + \tan^2{\theta})

    0 = -\left(\Delta y + \dfrac{g \Delta x^2}{2v_0^2}\right) + \Delta x \tan{\theta} - \dfrac{g \Delta x^2}{2v_0^2}\tan^2{\theta}

    a = - \dfrac{g \Delta x^2}{2v_0^2}

    b = \Delta x

    c = -\left(\Delta y + \dfrac{g \Delta x^2}{2v_0^2}\right)

    use the quadratic formula to solve for \tan{\theta} ... you will get two valid solutions.

    inverse tangent yields \theta \approx 60^\circ and \theta \approx 10^\circ
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    Anyway, the problem remains tough to solve...

    After simplifying we get:

    -1.42884 + 2.7\sin\theta\cos\theta + \cos^2\theta = 0

    Spoiler:
    http://www.wolframalpha.com/input/?i=-1.42884+%2B+2.7sin(x)cos(x)+%2B+cos^2(x)+%3D+0

    And putting n = 0, the smallest angles are 1.04 rad = 59.7 degrees and 0.173 rad = 9.9 degrees


    EDIT: How can I forget 1/cos^2A = sec^2A = 1 + tan^2A
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Sep 2010
    From
    Oslo
    Posts
    57
    Thanks
    4
    I can see you already posted the answer Unknown (edit: and Skeeter), but hey I needed to practice my LaTex skills anyways, so here it goes:

    - \dfrac12 g(\dfrac{2.7}{5\cos\theta})^2+5 \sin\theta (\dfrac{2.7}{5\cos\theta})+1=0

    <br />
(-\frac{1}{2}(2.7)^2(\frac{1}{5^2})\frac{1}{cos^2\th  eta}+2.7tan\theta+1=0

    -1.43(sec^2\theta)+2.7tan\theta+1=0

    -1.43(1+tan^2\theta)+2.7tan\theta+1=0

    -1.43tan^2\theta+2.7tan\theta-0.43=0


    Using the quadratic formula I get:

    tan\theta=1.71 and tan\theta=0.176

    Which gives

    \theta=59.7^\circ

    and

    \theta=10^\circ
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 9th 2010, 08:11 PM
  2. Need help with algebra based physics problem
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: May 6th 2010, 09:21 AM
  3. help w/ a algebra physics problem
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 22nd 2010, 10:46 PM
  4. Algebra/Physics-like problem?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 13th 2009, 02:48 PM
  5. physics, acceleration, physics problem
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: September 29th 2007, 04:50 AM

Search Tags


/mathhelpforum @mathhelpforum