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

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    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. )
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    someone please help! I really do not know how to solve this
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    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
    x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2
    and
    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 \frac{d^2x}{dt^2} = a_x, etc.)

    We need suitable initial conditions, of course. First off a_x = 0. Also the point (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 (x_0, y_0) = (0, 1.5~m). That way none of my y values are negative. We need to recognize the v_{0x} and v_{0y} as initial velocity components, so v_{0x} = 100 \cdot cos(30)~m/s and v_{0y} = 100 \cdot sin(30)~m/s.

    So the parametric equations of motion are:
    x = 100t~cos(30)
    and
    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 \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 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:
    x = 100t~cos(30)
    and
    0 = 1.5 + 100t~sin(30) - 4.9t^2
    for x.

    -Dan
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