physics

A soccer ball is kicked with an initial speed of 10.8 m/s in a direction 24.9° above the horizontal. Calculate the magnitude of its velocity 0.310 s after being kicked. What was its direction? Calculate the magnitude of its velocity 0.620 s after being kicked. What was its angle relative to the horizontal (choose positive for above and negative for below)?

asked by Anonymous
  1. Vo = 10.8m/s[24.9o]
    Xo = 10,8*cos24.9 = 9.8 m/s
    Yo = 10.8*sin24.9 = 4.55 m/s

    a. Y = Yo + g*t
    Y = 4.55 + (-9.8)*0.310 = 1.51 m/s.

    V^2 = Xo^2 + Y^2
    V^2 = 9.8^2 + 1.51^2 = 98.3
    V = 9.9 m/s.

    b. = tan A = Y/Xo = 1.51/9.8 = 0.15408
    A = 8.76o = The direction.

    c. Y = Yo + g*t
    Y = 4.55 + (-9.8)*0.620 = -1.53 m/s.

    V^2 = Xo^2 + Y^2
    V^2 = 9.8^2 + (-1.53^2) = 98.38
    V = 9.9 m/s.

    tan B = Y/Xo = -1.53/9.8 = -0.15612
    B = -8.87o = The direction.

    NOTE: The negative velocity(-1.53) means that the ball is falling and each
    point during the rise time is repeated during the fall time.

    posted by Henry

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