射彈是涉及二維的運動。為瞭解決彈丸運動問題，取兩個相互垂直的方向（通常，我們使用“水平”和“垂直”方向），並將所有向量（位移、速度、加速度）作為沿每個方向的分量。在彈丸中，垂直運動獨立於水平運動。所以，運動方程可以分別適用於水平運動和垂直運動。

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.

當以一定角度拋射的彈丸達到最大高度時，其垂直速度分量為0；當彈丸達到拋射時的水平時，其垂直位移為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.

在做下麵的計算時，我們在垂直方向上取向上的向量為正，在水平方向上取向右的向量為正。

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.

嚴格地說，由於空氣阻力，路徑不是拋物線。相反，形狀變得更“擠壓”，粒子的範圍變小。

最初，由於地球試圖吸引物體向下，物體的垂直速度在下降。最終，垂直速度達到0。物體現在已經達到最大高度。然後，物體開始向下移動，其向下速度隨著物體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

一個人站在30米高的樓頂，以15米s-1的速度從樓頂水平拋下一塊石頭。找到

a） 物體到達地面所用的時間，

b） 離它降落的大樓有多遠

c） 物體到達地面時的速度。

物體的水平速度是不變的，因此它本身在計算時間時是沒有用的。我們知道物體從樓頂到地面的垂直位移。如果我們能找出物體到達地面所花的時間，我們就能知道物體在這段時間內水平移動的幅度。

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的速度從地面踢起，與地面成20度角。假設沒有空氣阻力，找出球落地的距離。

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.