Results 1 to 3 of 3

Thread: Help!World Problem:Calculus III Projectile Motion (Velocity & Acceleration)

  1. #1
    Junior Member
    Joined
    May 2007
    Posts
    69

    Help!World Problem:Calculus III Projectile Motion (Velocity & Acceleration)

    Please someone help yikes!

    Assuming there is no resistance and knowing that a(t)= -9.8 m/s^2

    Determine the maximum height and range of a projectile fired at a height of 1.5 meters above the ground with an initial velocity of a 100 meters per second and at an angle of 30 degrees above the horizontal.

    Answers in back of book:
    maximum height: 129.1 meters
    range: 886.3 meters
    (how do you go about solving this problem using Vector-Valued Functions such as r(t)= x(t)i + y(t)j etc. )
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    May 2007
    Posts
    69
    someone please help! I really do not know how to solve this
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,152
    Thanks
    731
    Awards
    1
    Quote Originally Posted by googoogaga View Post
    Please someone help yikes!

    Assuming there is no resistance and knowing that a(t)= -9.8 m/s^2

    Determine the maximum height and range of a projectile fired at a height of 1.5 meters above the ground with an initial velocity of a 100 meters per second and at an angle of 30 degrees above the horizontal.

    Answers in back of book:
    maximum height: 129.1 meters
    range: 886.3 meters
    (how do you go about solving this problem using Vector-Valued Functions such as r(t)= x(t)i + y(t)j etc. )
    The equation of motion can be expressed parametrically as you state with
    $\displaystyle x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2$
    and
    $\displaystyle y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2$

    (You can derive these from knowing that the acceleration components in both x and y directions are constant. Then $\displaystyle \frac{d^2x}{dt^2} = a_x$, etc.)

    We need suitable initial conditions, of course. First off $\displaystyle a_x = 0$. Also the point $\displaystyle (x_0, y_0)$ acts as an origin for the problem. You can select any one that you like; for my purposes I'm going to choose $\displaystyle (x_0, y_0) = (0, 1.5~m)$. That way none of my y values are negative. We need to recognize the $\displaystyle v_{0x}$ and $\displaystyle v_{0y}$ as initial velocity components, so $\displaystyle v_{0x} = 100 \cdot cos(30)~m/s$ and $\displaystyle v_{0y} = 100 \cdot sin(30)~m/s$.

    So the parametric equations of motion are:
    $\displaystyle x = 100t~cos(30)$
    and
    $\displaystyle y = 1.5 + 100t~sin(30) - 4.9t^2$

    We want the maximum height for the first problem. What is the condition for a moving object to be at an extremum? When the first derivative is 0, of course. So when $\displaystyle \frac{dy}{dt} = 0$ the particle is at its maximum height. As per my usual, I'll leave it to you to prove that this is indeed a maximum. (Remember you want y(t) not just t.)

    The second problem asks about the range. This is the x value (actually $\displaystyle x - x_0$ value) for when the projectile is at ground level, that is to say y = 0 in my coordinate system. So you need to solve the system of equations:
    $\displaystyle x = 100t~cos(30)$
    and
    $\displaystyle 0 = 1.5 + 100t~sin(30) - 4.9t^2$
    for x.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Dec 29th 2009, 07:25 AM
  2. projectile motion problem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Apr 27th 2009, 08:29 PM
  3. Replies: 1
    Last Post: Feb 19th 2009, 02:30 PM
  4. Projectile Motion Problem
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: Nov 24th 2008, 02:53 PM
  5. projectile motion problem
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: Apr 12th 2007, 07:47 AM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum