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Math Help - Antiderivatives/derivatives

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    Antiderivatives/derivatives

    I do not know what to do with the 45 degrees!

    A ball is shot at an angle of 45 degrees into the air with initial velocity of ft/sec. Assuming no air resistance, how high does it go?


    How far away does it land?
    Hint: The acceleration due to gravity is 32 ft per second squared.
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    Quote Originally Posted by MathNeedy18 View Post
    I do not know what to do with the 45 degrees!

    A ball is shot at an angle of 45 degrees into the air with initial velocity of ft/sec. Assuming no air resistance, how high does it go?


    How far away does it land?
    Hint: The acceleration due to gravity is 32 ft per second squared.
    I only have a moment, so I'll be brief.

    The motion in perpendicular directions is independent of each other. So we have:
    x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2

    y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2

    Here a_x = 0 because nothing is making it accelerate in the x direction and a_y = -g.

    Now, you have an initial velocity v_0 pointed at 45 degrees above the horizontal. This is a vector. So your components in the x and y direction will depend on this angle.

    For example, let's call +y upward and +x along the ground in the relative direction the ball was shot in. Then
    v_{0x} = v_0~cos(45^o)
    and
    v_{0y} = v_0~sin(45^o)

    -Dan
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