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Two-Dimensional Kinematics, Projectile

  1. Apr 11, 2005 #1
    Q. At t = 0, an object is projected with a speed v0 = 30 m/s at an angle q0 = 30° above the horizontal. The axes on the diagram show the x and y directions that are to be considered positive.

    For parts a-g, calculate the requested quantities at t = 10 s into the flight. ( Use 9.81 m/sec2 for g. )


    --------------------------------------------------------------------------------
    a) The vertical acceleration of the object:
    ay = m/s2 *

    HELP: What force is acting on the object in the vertical direction?
    HELP: Is it moving faster or slower as it travels upward?


    --------------------------------------------------------------------------------
    b) Its horizontal acceleration:
    ax = m/s2 *
    0 OK


    --------------------------------------------------------------------------------
    c) Its vertical velocity:
    vy = m/s *
    -83.1 OK

    HELP: Use the definition of acceleration.


    --------------------------------------------------------------------------------
    d) Its horizontal velocity:
    vx = m/s *
    25.98 OK


    --------------------------------------------------------------------------------
    e) The angle to the horizontal at which the object is traveling (an angle above the horizontal should be reported as a positive number; an angle below the horizontal should be reported as a negative number). Please give your answer in degrees:
    angle = ° *
    -72.59 OK


    --------------------------------------------------------------------------------
    f) Its vertical displacement, from where it started:
    y = m *
    -340.5 OK

    HELP: Look at the equations in your book.
    HELP: Look for an equation relating initial velocity, change in position, acceleration, and time. Use your previous answers.


    --------------------------------------------------------------------------------
    g) Its horizontal displacement, from where it started:
    x = m *
    259.8 OK


    --------------------------------------------------------------------------------
    h) At what time does the object reach its maximum height?
    ty, max = s
    2*83.1*sin(72.59)/9.81 NO


    I just need some help with part h.

    At the max, height final velocity is zero, I know that

    I tried using t = v_y/g

    and plugging this into

    y-yo = v0sintheta*t - 1/2 g* t^2

    didn't work!

    Pl. Help
     

    Attached Files:

  2. jcsd
  3. Apr 11, 2005 #2
    I tried using t = v_y/g isnt right.

    For an object under constant acceleration with no initial velocity it takes

    [tex] t = \sqrt{\frac{2d}{g}}[/tex]
     
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