射弹是涉及二维的运动。为了解决弹丸运动问题，取两个相互垂直的方向（通常，我们使用“水平”和“垂直”方向），并将所有矢量（位移、速度、加速度）作为沿每个方向的分量。在弹丸中，垂直运动独立于水平运动。所以，运动方程可以分别适用于水平运动和垂直运动。

To solve projectile motion problems for situati*** where objects are thrown on Earth, the acceleration due to gravity,  , is always acting vertically downward. If we neglect effects of air resistance, then the horizontal acceleration is 0. In this case, the horizontal component of the projectile’s velocity remains unchanged.

当以一定角度抛射的弹丸达到最大高度时，其垂直速度分量为0；当弹丸达到抛射时的水平时，其垂直位移为0。

On the diagram above, I have shown some typical quantities you should know in order to solve projectile motion problems.  is the initial velocity and , is the final velocity. The subscripts and refer to the horizontal and vertical components of these velocities, separately.

在做下面的计算时，我们在垂直方向上取向上的向量为正，在水平方向上取向右的向量为正。

Let us c***ider the vertical displacement of the particle with time. The initial vertical velocity is . At a given time, the vertical displacement , is given by . If we are to draw a graph of vs. , we find that the graph is a parabola because has a dependence on . i.e., the path taken by the object is a parabolic one.

严格地说，由于空气阻力，路径不是抛物线。相反，形状变得更“挤压”，粒子的范围变小。

最初，由于地球试图吸引物体向下，物体的垂直速度在下降。最终，垂直速度达到0。物体现在已经达到最大高度。然后，物体开始向下移动，其向下速度随着物体被重力向下加速而增加。

For an object thrown from the ground at speed , let’s try to find the time taken for the object to reach the top. To do this, let’s c***ider the motion of the ball from when it was thrown to when it reaches the maximum height.

The vertical component of the initial velocity is . When the object reaches the top, the object’s vertical velocity is 0. i.e. . According to the equation , the time taken to reach the top = .

If there is no air resistance, then we have a symmetrical situation, where the time taken for the object to reach the ground from its maximum height is equal to the time taken by the object to reach the maximum height from the ground in the first place. The total time that the object spends in air is then, .

If we c***ider the object’s horizontal motion, we can find the object’s range. This is the total distance traveled by the object before it lands on the ground. Horizontally, becomes (because horizontal acceleration is 0). Substituting for , we have: .

例1

一个人站在30米高的楼顶，以15米s-1的速度从楼顶水平抛下一块石头。找到

a） 物体到达地面所用的时间，

b） 离它降落的大楼有多远

c） 物体到达地面时的速度。

物体的水平速度是不变的，因此它本身在计算时间时是没有用的。我们知道物体从楼顶到地面的垂直位移。如果我们能找出物体到达地面所花的时间，我们就能知道物体在这段时间内水平移动的幅度。

So, let us start with the vertical motion from when it was thrown to when it reaches the ground. The object is thrown horizontally, so the initial vertical velocity of the object is 0. The object would experience a c***tant vertical acceleration downwards, so m s-2. The vertical displacement for the object is m. Now we use , with . So, .

To solve part b) we use horizontal motion. Here, we have 15 m s-1, 6.12 s, and 0. Because horizontal acceleration is 0, the equation becomes or, . This is how much farther from the building the object would land.

To solve part c) we need to know the final vertical and horizontal velocities. We already know the final horizontal velocity, m s-1. We need to again c***ider the vertical motion to know the object’s final vertical velocity, . We know that , -30 m and m s-2. Now we use , giving us . Then, . Now we have the horizontal and vertical components of the final speed. The final speed is, then, m s-1.

例2

足球以25米s-1的速度从地面踢起，与地面成20度角。假设没有空气阻力，找出球落地的距离。

This time, we have a vertical component for initial velocity too. This is,  m s-1. The initial horizontal velocity is  m s-1.

When the ball lands, it comes back to the same vertical level. So we can use  , with  . This gives us . Solving the quadratic equation, we get a time of  0 s or 1.74 s. Since we are looking for the time when the ball lands, we take  1.74 s.

Horizontally, there is no acceleration. So we can substitute the time of the ball’s landing into the horizontal equation of motion:  m. This is how far away the ball will land.