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  • 天同
    Lv 7
    7 年 前
    最佳解答

    3. Yes, you are right. The kinetic energy on jumping upward needs to be included.

    4. (a) Observer on ground: the ball moves forward and falls in a parabolic trajectory.

    Observer on toy are: the ball falls backward following a (nearly) parabolic trajectory.

    (b)(i) acceleration of toy car = 10/4 m/s^2 = 2.5 m/s^2

    Use v^2 = u^2 + 2a.s to find speed of toy car after travelling 2 m

    v^2 = 0^2 + 2 x 2.5 x 2 (m/s)^2

    v^2 = 10 (m/s)^2

    v = 3.162 m/s

    Consider the vertical motion of the ball after it is released,

    use: s = ut + (1/2)at^2

    with s = 0.5 m, u = 0 m/s, a = -g(=-10 m/s^2), t=?

    0.5 = (1/2).(10)t^2

    t = 0.3162 s

    It takes 0.3162 s for the ball to land on the toy car

    Use: s = ut + (1/2)at^2 to find the distance travelled by the toy car in 0.3162 s

    hence, s = [3.162 x 0.3162 + (1/2) x 2.5 x 0.3162^2] m = 1.125 m

    (ii) Horizontal distance travelled by the falling ball

    = 3.162 x 0.3162 m = 1 m

    Hence, s = (1.125 - 1) m = 0.125 m

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