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? 發問於 科學及數學其他 - 科學 · 1 十年前

physics problems

1.an outfielder throws a baseball to the first baseman at a peed of 19.6m/s and an angle of 30 degrees above the horizontal.

if the ball is caught at the same height from which it was thrown, calculate the amount of time the ball was in the air.

2.a motor used 120 watts of power to raise a 15N object in 5s. through what vertical distance was the object raised?

explain in details please!!!!

2 個解答

評分
  • 1 十年前
    最愛解答

    (1) The initial velocity of the baseball can be divided into vertical and horizontal components:

    Since the ball is caught at the same height from which it was thrown, we can say that its displacement in the vertical direction is zero at the moment it was caught.

    Now, vertical component of its initial velocity is given by:

    v = 19.6 sin 30(deg)

    = 9.8 m/s

    Taking g = 9.8 m/s2 where it is the acceleration of free fall due to gravity and applying the equation:

    s = ut + (1/2)at2 in the vertical direction,

    Sub s = 0, u = 9.8 and a = -9.8 (since it is downward), we have:

    0 = 9.8t - 4.9t2

    t = 0 (rejected) or t = 2

    So the duration of the baseball when it was in the air is 2 seconds.

    (2) Assuming that the motor is 100% efficient, the work done on the object is 120 x 5 = 600 J which eventually becomes the gravitational p.e. of the object as it was raised vertically.

    Hence by gravitational p.e. = mgh, where mg is equivalent to the object's weight, i.e. 15N and h is the height through which the object was raised.

    600 = 15h

    h = 40m.

    資料來源: My Physics knowledge
  • Karin
    Lv 6
    1 十年前

    1.an outfielder throws a baseball to the first baseman at a peed of 19.6m/s and an angle of 30 degrees above the horizontal.

    if the ball is caught at the same height from which it was thrown, calculate the amount of time the ball was in the air.

    By s = ut-(1/2)gt^2, assuming g = 10m/s^2

    0 = (19.6sin30)t -(1/2)(10)t^2

    t = 9.8s

    2.a motor used 120 watts of power to raise a 15N object in 5s. through what vertical distance was the object raised?

    By Power = Work Done / Time,

    120 = 15xdistance/5

    distance = 40m

    2007-03-28 09:09:39 補充:

    Correction for question, 0 = (19.6sin30)t -(1/2)(10)t^20 = 19.6(1/2) - (1/2)(10)tt = 19.6/10t = 1.96second

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