**Projectile motion**

A **projectile** is an object upon which the only force acting is **gravity.** There are a variety of examples of **projectiles**. An object dropped from rest is a **projectile** (provided that the influence of air resistance is negligible). From here we can conclude that free fall is a type of projectile. An object that is thrown vertically upward is also a projectile (provided that the influence of air resistance is negligible). And an object which is thrown upward at an angle to the horizontal is also a projectile (provided that the influence of air resistance is negligible). A projectile is any object that once projected or* *dropped continues in motion by its own inertia and is influenced only by the downward force of gravity. The motion of an object when it is thrown and falls to the ground is called as **projectile motion**. The path the object follows is a parabolic path. Some types of projectiles are shown below:

Here we will focus more on the **motion in 2 dimensions** (i.e. when an object is thrown upwards at an angle to the horizontal direction), because when one understands this type of **projectile**, the other types can be easily understood. It is convenient to perform the study of this motion according the **horizontal** and **vertical direction** separately. As mentioned before, the only force acting on the object is the **gravitational force**, therefore the only acceleration acting on the object is the **acceleration caused by gravity** (gravitational acceleration). Thus, with respect to the horizontal direction, there is no force actiing on the object and therefore no acceleration does exist in this component of motion. Hence, the **horizontal component** of projectile motion is a **uniform motion** and the equation used here is the **equation of the uniform motion**.

Thus, if we denote the horizontal direction as **x-direction** and the vertical direction as **y-direction**, we obtain 2 types of motions:

1. Horizontally (according to x-direction) we have a **uniform motion**

2. Vertically (according to y-direction) we have a **uniformly deccelerated motion when going up** (as the gravitational acceleration acts downwards) and **uniformly accelerated motion when falling down** ( we have seen it when studying the free fall topic)

When objects are thrown upwards at an angle \((\alpha)\) to the horizontal direction, its initial velocity \((\vec{v_0})\) is divided into 2 components:

1. Horizontal component of the initial velocity \(\vec{v_0x}=\vec{v_0}\times{\cos{\alpha}}\)

2. Vertical component of the initial velocity \(\vec{v_0y}=\vec{v_0}\times\sin\alpha\)

By using the abovementioned notations, we obtain the following 4 equations for the projectile motion:

Horizontally: \(\vec{\Delta{x}}=\vec{v_0}_x\times{t}=\vec{v_0}\cos\alpha\times{t}\)

Vertically: 1. \(\vec{v}_y=\vec{v_0}_y+\vec{g}\times{t}=\vec{v_0}\sin{\alpha}+\vec{g}\times{t} \)

2. \(\vec{h}=\vec{v_0}_y\times{t}+\frac{\vec{g}\times{t}^2}{2}=\vec{v_0}\sin{\alpha}\times{t}+\frac{\vec{g}\times{t}^2}{2} \)

3. \(\vec{v}_y^2-\vec{v_0}_y^2=2\times \vec{g}\times{\vec{h}} \) or \(\vec{v_y}^2 - \vec{(v_0}\sin{\alpha})^2=2\times{\vec{g}}\times{\vec{h}}\)

In order to make a complete study of projectile motion one has to determine the **position** of objects, their **velocity** and its **direction** at every instant.

The diagram below gives a more comprehensive view of the **most complicated** type of **projectile motion**; i.e. the **motion in two dimensions**

(Refer to the above diagram to gain a more comprehensive understanding of the following example)

__ Example 1:__ An object is thrown upwards by 40 m/s at an angle of 37

^{o}to the horizontal direction (cos 37

^{o}=0.8; sin 37

^{o}=0.6). Calculate:

a) The time of motion

b) The maximum height the object can reach

c) Its horizontal displacement until it falls on the ground

d) Its velocity just before touching the ground

(Take g=10 m/s^{2})

__Solution:__

**a)** To calculate the time we must refer to the vertical component of motion. First, we find the time of moving upwards up to the maximal height. At this height (as you can see in the figure) the vertical component of velocity \(\vec{v_y}\) is zero (because the object is moving only horizontally at this point. Thus,

\(\vec{v_y}=\vec{v_0}_y+\vec{g}\times {t_u}_p =\vec{v_0}\times\sin{\alpha}+\vec{g}\times {t_u}_p=0\)

\(0=40\times0.6+(-10)\times{t_u}_p\)

(Remember that the object is slowing down while going up, therefore the gravitational acceleration will be negative). Thus,

\(10\times{t_u}_p=24\)

\(t_{up} = 2.4\;s\)

*As the object will fall again to the original height after its flight, the time of raising up will be equal to the time of falling down. Therefore, \(t_{total}=2\times{t_u}_p=2\times{2.4\;s}=4.8\;s\)*

**b)** The maximal height can also be calculated by using the vertical components of this projectile motion. For this, we can use either the equation 2 or 3 of the vertical motion. Let's use for example the equation 3. In this equation, as we have previuosly mentioned, \(\vec{v_y}=0\) ;\(\vec{v_0}_y=\vec{v_0}sin{\alpha}=24\;m/s\). Also, during the upwards motion, \(\vec{g}=-10\;m/s^2\). Hence,

\(\vec{v}^2-\vec{v_0^2}=2\times{\vec{g}}\times{\vec{h}}_m\)

\(0^2-24^2=2\times({-10})\times{h_m}\)

\(-576=-20\times{h_m}\)

\(h_m=(-576)\div({-20})=28.8\;m\)

**c)** For the horizontal displacement \((\Delta{\vec{x}})\) we must first calculate the horizontal component of the initial velocity. Thus,

\(\vec{v_0}_x=\vec{v_0}cos\alpha=40\;m/s\times0.8=32\;m/s\)

Hence, \(\Delta{\vec{x}}=\vec{v_0}cos{\alpha}\times{t}=32\;m/s\times4.8\;s=153.6\;m\)

d) As for the final velocity, we only need to find the vertical component of final velocity, because its horizontal component remains always the same (32 m/s). Thus, taking only the second part of motion (falling down part) we have the initial vertical velocity of falling part equal to zero, because - as we have seen before - the motion in the highest point of projectile is only horizontal. Furthermore, the gravitational acceleration (g) is positive at this part because the object is falling down. Hence,

\(\vec{v_y}=\vec{v_m}_y+\vec{g}\times{t}_d\)

where \(\vec{v_m}_y\) is the vertical velocity at the highest point (=0) and \(t_d\) is the time of falling down (equal to 2.4 s). Thus,

\(\vec{v_y}=0+10\;m/s^2\times2.4\;s=24\;m/s\)

To calculate the magnitude of the final velocity, \((v_f)\) we must use the Pythagorean theorem. Hence,

\(\vec{v_f}^2=\vec{v_x}^2+\vec{v_y}^2=32^2+24^2=1024+576=1600\)

Thus, \(\vec{v_f}=\sqrt{1600}=40\;m/s\)

This value is the same as the value of the initial velocity (only the direction changes). Therefore, it is a confirmation of the fact that projectile motion is symmetrical to the vertical line passing through the maximal height. The projectile's trajectory is a **parabola**. For this reason, this type of motion is often called "**parabolic motion**".

** Example 2: **An object is thrown vertically upwards at 30 m/s. When will the object be 40 m above the ground? (Refer to the second part of the first figure shown at the beginning of this lesson)

First, we need to make an analysis by taking into consideration all the possibile situations regarding this motion.

1. The object may not reach the height 40 m and therefore we will not be able to find a result for the time required

2. The maximal height of the object may be 40 m. In such a case, there is only a value for the time required.

3. The object may go higher than 40 m. In such a case, we will obtain 2 values for the time; one while going up and one while falling down after having reached the highest point.

In our example, \(\vec{v_0}=\vec{v_0}_y=30\;m/s\) and \(\vec{g}=-10\;m/s^2\). Also \(\vec{h}=40\;m.\)

Using these clues in the second equation of vertical motion, we get:

\(\vec{h}=\vec{v_0}_y\times{t}+\frac{\vec{g}\times{t}^2}{2}\)

\(40=30\times{t}+\frac{({-10})\times{t}^2}{2}\)

\(40=30\times{t}-5\times{t}^2\)

Simplifying by 5 and arranging the equation we obtain:

\(t^2-6t+8=0\)

\(\Delta=b^2-4ac=(-6)^2-4\times{1}\times{8}=36-32=4\)

From here we understand that we will obtain 2 different values for **t **(case 3). Therefore,

\(t_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{6-2}{2}=2\;s\)

\(t_2=\frac{-b+\sqrt{\Delta}}{2a}=\frac{6+2}{2}=4\;s\)

Thus, the object will be twice at the height 40 m: one while moving upwards (after 2 s) and one while falling downwards (after 4 s).