2.7 由于重力而加速

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  由于重力加速

  

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  In the absence of air resistance, all objects fall toward the Earth with the same . Parachutists, like the one from the U.S. Army Parachute Team shown above, make maximum use of air resistance in order to limit the acceleration of the fall.
  ::在没有空气抵抗的情况下,所有物体都以同样的降落伞手向地球倾斜。像上面显示的美军降落伞队那样,最大限度地利用空气抵抗来限制坠落速度的加速。

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  Acceleration Due to Gravity
  ::由于重力加速

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  One of the most common examples of uniformly accelerated motion is that an object allowed to drop will fall vertically to the Earth due to gravity. In treating falling objects as uniformly accelerated motion, we must ignore air resistance. Galileo’s original statement about the motion of falling objects is:
  ::统一加速运动的最常见例子之一是,允许坠落的物体会由于重力而垂直坠落到地球。 在将坠落的物体作为统一加速运动对待时,我们必须忽视空气抵抗。 伽利略最初关于坠落物体运动的说法是:

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  At a given location on the Earth and in the absence of air resistance, all objects fall with the same .
  ::在地球上某个特定地点,在没有空气阻力的情况下,所有物体都处于同一状态。

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  We call this acceleration due to gravity on the Earth and we give it the symbol $\begin{array}{r}g\end{array}$ . The value of $\begin{array}{r}g\end{array}$ is 9.81 m/s ² in the downward direction. All of the equations involving constant acceleration can be used for falling bodies but we insert  $\begin{array}{r}g\end{array}$ wherever “ $\begin{array}{r}a\end{array}$ ” appeared and the value of  $\begin{array}{r}g\end{array}$ is always 9.81 m/s ² .
  ::我们称这种加速因地球重力而加速,我们给它符号g。g值向下为9.81 m/s2。所有涉及恒定加速度的方程式都可以用于下降的物体,但我们在出现“a”时插入g值,而g值总是9.81 m/s2。

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  In the absence of air resistance, is the cliff diver’s acceleration 9.81 m/s ²  in the simulation below? How do you know? What information can you derive from the graphs?
  ::在没有空气抵抗的情况下,悬崖潜水员的加速度在下面的模拟中是否为9.81米/秒?你怎么知道?从图表中可以得到什么信息?

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  Example:  A rock is dropped from a tower 70.0 m high. How far will the rock have fallen after 1.00 s, 2.00 s, and 3.00 s? Assume the is positive downward.
  ::示例: 岩石从70.0米高的塔上坠落。 在1秒、 2.00秒和 3.00秒之后, 岩石会跌落到多远? 假设是正向下跌 。

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  Solution:  We are looking for displacement and we have time and acceleration. Therefore, we can use $\begin{array}{r}d=\frac{1}{2}a{t}^2\end{array}$ .
  ::解决方案:我们正在寻找流离失所,我们有时间和加速。因此,我们可以使用 d=12at2。

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  Displacement after 1.00 s:  $\begin{array}{r}\left(\frac{1}{2}\right)(9.81{\text{m/s}}^2)(1.00\text{s}{)}^2=4.91\text{m}\end{array}$
  ::1.00秒后流离失所12)(981米/秒)(1.00秒)2=4.91米

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  Displacement after 2.00 s:  $\begin{array}{r}\left(\frac{1}{2}\right)(9.81{\text{m/s}}^2)(2.00\text{s}{)}^2=19.6\text{m}\end{array}$
  ::2.00秒后流离失所12)(9.81米/秒)(2.00秒)2=19.6米

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  Displacement after 3.00 s:  $\begin{array}{r}\left(\frac{1}{2}\right)(9.81{\text{m/s}}^2)(3.00\text{s}{)}^2=44.1\text{m}\end{array}$
  ::3.00秒后流离失所12)(9.81米/秒)(3.00秒)2=44.1米

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  Example:  (a) A person throws a ball upward into the air with an initial of 15.0 m/s. How high will it go before it comes to rest? (b) How long will the ball be in the air before it returns to the person’s hand?
  ::实例a) 一个人向上投球,首字母为15.0米/秒。球在休息前会升到多高? (b) 球在空气中多久才能回到人的手中?

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  Solution:  In part (a), we know the initial velocity (15.0 m/s), the final velocity (0 m/s), and the acceleration (-9.81 m/s ² ). We wish to solve for the displacement, so we can use  $\begin{array}{r}v{{}_f}^2=v{{}_i}^2+2ad\end{array}$ and solve for $\begin{array}{r}d\end{array}$ .
  ::解答:在(a)部分,我们知道初始速度(1.5万米/秒)、最终速度(0.0米/秒)和加速度(9.81米/秒),我们希望解决移位问题,这样我们就可以使用 vf2=vi2+2ad 并解决 d。

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  $\begin{array}{r}d=\frac{v{{}_f}^2-v{{}_i}^2}{2a}=\frac{(0\text{m/s}{)}^2-(15.0\text{m/s}{)}^2}{(2)(-9.81{\text{m/s}}^2)}=11.5\text{m}\end{array}$
  ::d=vf2-vi22a=(0.m/s)2-(5.0m/s)2(2)-9.81m/s2=11.5m

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  There are a number of methods by which we can solve part (b). Probably the easiest is to divide the distance traveled by the to get the time going up and then double this number since the motion is symmetrical—that is, time going up equals the time going down.
  ::我们可以用多种方法来解决(b)部分。也许最容易的方法就是将时间上升所走的距离分开,然后将这个数字翻一番,因为运动是对称的,也就是说,时间上升等于时间下降。

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  The average velocity is half of 15.0 m/s, or 7.5 m/s, and dividing this into the distance of 11.5 m yields 1.53 s. This is the time required for the ball to go up and the time for the ball to come down will also be 1.53 s, so the total time for the trip up and down is 3.06 s.
  ::平均速度为15.0米/秒半,或7.5米/秒,将其分为11.5米的距离等于1.53秒。这是球上升和球下降所需的时间,也是1.53秒,因此,起降总时间为3.06秒。

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  Example:  A stone is dropped from the top of a cliff. It is hits the ground after 5.5 s. How high is the cliff?
  ::示例: 一块石头从悬崖顶上掉下来。 它在5. 5 秒后撞击到地面。 悬崖有多高?

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  Solution: 
  ::解决方案 :

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  $\begin{array}{r}d=v_{i}t+\frac{1}{2}a{t}^2=(0\text{m/s})(5.5\text{s})+\left(\frac{1}{2}\right)(9.81{\text{m/s}}^2)(5.5\text{s}{)}^2=150\text{m}\end{array}$
  ::d=vit+12at2=(0 m/s)(5.5 s)+(12) (9.81 m/s2(5.5 s)2=150 m

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  Further explore the acceleration of a model rocket due to gravity in the simulation below. Can you use the graphs to determine when the rocket is speeding up or slowing down?
  ::在下面的模拟中进一步探索模型火箭因重力而加速的速度。 您能否用图表来确定火箭在何时加速或减速?

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  Summary
  ::摘要

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  • At any given location on the Earth and in the absence of air resistance, all objects fall with the same uniform acceleration.
    ::在地球上任何特定地点,在没有空气阻力的情况下,所有物体都以相同的加速度坠落。
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  • We call this acceleration the acceleration due to gravity on the Earth and we give it the symbol $\begin{array}{r}g\end{array}$ . 
    ::我们称这种加速因地球重力而加速 我们给它符号g
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  • The value of  $\begin{array}{r}g\end{array}$ is 9.81 m/s ² .
    ::g 值为 9.81 m/s2。

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  Review
  ::回顾

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  1. A baseball is thrown vertically into the air with a speed of 24.7 m/s.
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     1. How high does it go? 
        ::它能飞多高?
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     2. How long does the round trip up and down require?
        ::往返旅行需要多久?
     
     ::棒球以24.7米/秒的速度垂直投向空气中。 速度有多高? 往返需要多久?
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  2. A salmon jumps up a waterfall 2.4 m high. With what minimum speed did the salmon leave the water below to reach the top?
     ::鲑鱼以什么最低速度 将下方的水排到顶端?
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  3. A kangaroo jumps to a vertical height of 2.8 m. How long will it be in the air before returning to Earth?
     ::袋鼠跳到2.8米的垂直高度。 在返回地球之前,它会在空气中停留多久?

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  Explore More
  ::探索更多

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  This video offers a discussion and demonstration of the acceleration due to gravity.
  ::这段录像讨论并展示了重力加速的情况。

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  1. What is the gravitational acceleration given in the video? Why does it differ from that given in this text?
     ::视频中给出的引力加速度是什么?为什么它不同于本文中给出的加速度?
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  2. Why does the ball travel further in later time intervals than in the earlier ones?
     ::为什么球在较晚的时间间隔比早先的时间间隔更远?