# How should I define the projectile movement

## Projectile motion (physics): definition, equations, problems (with examples)

Let me introduce you to bury a cannon to tear down the walls of an enemy castle so that your army can tower and win the victory. If you like how fast I move the ball,

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Imagine occupying a cannon to tear down the walls of an enemy castle so your army can storm and achieve victory. If you know how fast the ball is moving when it leaves the cannon and how far away the walls are, what starting angle do you need to fire the cannon to successfully hit the walls?

This is an example of a projectile motion problem, and you can solve this and many similar problems using the constant acceleration equations of kinematics and some basic algebra.

Projectile movement Physicists describe two-dimensional movements in which the only acceleration that the object in question experiences is the constant downward acceleration due to gravity.

The constant acceleration on the earth's surface a corresponds to G = 9.8 m / s2and an object that is projectile moving is in freefall with this as the only source of acceleration. In most cases, it takes the path of a parabola, so the motion has both a horizontal and a vertical component. Fortunately, while this would have a (limited) effect in real life, most projectile movement problems in high school physics ignore the effect of drag.

You can solve projectile movement problems with the value of G and some other basic information about the current situation, such as the initial velocity of the projectile and the direction in which it is moving. Learning to solve these problems is essential to passing most introductory physics courses, and it introduces you to the key concepts and techniques that you will also need in later courses.

### Projectile motion equations

The projectile motion equations are the constant acceleration equations from Kinematics, since acceleration due to gravity is the only source of acceleration that you need to consider. The four main equations you will need to solve a projectile motion problem are:

v = v_0 + um s = bigg (frac {v + v_0} {2} bigg) t s = v_0t + frac {1} {2} um ^ 2 v ^ 2 = v_0 ^ 2 + 2as

Here, v stands for speed, v0 is the initial speed, a is the acceleration (which is equal to the downward acceleration of) G for all projectile movement problems), s is the shift (from the starting position) and as always you have time t.

These equations are technically only valid for one dimension and can actually be represented by vector quantities (including velocity) v, Initial speed v0 and so on), but in practice you can just use these versions separately, once in the X-Direction and once in the y- Direction (and if you've ever had a three-dimensional problem in the z-Direction too).

It is important to remember that this is the case is only used for constant accelerationThis makes them perfect for describing situations where the influence of gravity is the only acceleration, but unsuitable for many real-world situations where additional forces need to be taken into account.

In simple situations, this is all you need to describe the movement of an object. However, if needed, you can also consider other factors, such as: B. the height from which the projectile was shot, or even solve for the highest point of the projectile on its way.

### Fixed issues with projectile movement

Now that you've seen the four versions of the projectile motion formula that you must use to solve problems, it is time to think about the strategy that you will use to solve a projectile motion problem.

The basic approach is to split the problem into two parts: one for horizontal movement and one for vertical movement. This is technically referred to as the horizontal component and the vertical component, and each has a corresponding set of sizes, such as B. the horizontal speed, the vertical speed, the horizontal displacement, the vertical displacement and so on.

With this approach, you can use the kinematic equations and note this time t is the same for horizontal and vertical components, but things like initial speed have different components for initial vertical speed and initial horizontal speed.

The bottom line to understand is that for two-dimensional motion, any The angle of movement can be divided into a horizontal and a vertical component. In this case, however, there is a horizontal and a vertical version of the equation in question.

Neglecting the effects of drag massively simplifies the problems of projectile movement because in a projectile movement problem (free fall) the horizontal direction will never have any acceleration because the influence of gravity is only vertical (i.e. towards the surface of the earth).

This means that the horizontal component of speed is just a constant speed and movement will only stop when gravity brings the projectile to ground level. This can be used to determine the flight time as this is entirely dependent on y-Directional movement and can be calculated entirely based on vertical displacement (i.e. time) t When the vertical offset is zero, the flight time is displayed.

### Trigonometry for projectile movement problems

If the problem in question provides a starting angle and speed, then you need to use trigonometry to find the horizontal and vertical speed components. Once you do that, you can use the methods outlined in the previous section to actually solve the problem.

Essentially, you are creating a right triangle with the hypotenuse sloping at the starting angle (θ) and the magnitude of the speed as the length, and then the adjacent side is the horizontal component of the speed and the opposite side is the vertical speed.

Draw the right triangle as indicated and you will see that you can find the horizontal and vertical components based on the trigonometric identities:

{cos}; θ = frac {{adjacent}} {{hypotenuse}} {sin}; θ = frac {{opposite}} {{hypotenuse}}

So these can be rearranged (and with opposite = vy and adjacent = vXi.e. the vertical speed component or the horizontal speed component and hypotenuse = v0, the initial speed) to give:

v_x = v_0 cos (θ) v_y = v_0 sin (θ)

This is the trigonometry you need to do to fix projectile motion problems: inserting the starting angle into the equation, using the sine and cosine functions of your calculator, and multiplying the result by the initial velocity of the projectile.

To illustrate this with an example, the components at an initial speed of 20 m / s and a starting angle of 60 degrees are:

begin {align} v_x & = 20; {m / s} × cos (60) & = 10; {m / s} v_y & = 20; {m / s} x sin (60) & = 17.32; {m / s} end {align}

### Example of a projectile movement problem: An exploding fireworks display

Imagine a fireworks display has a fuse designed to explode at the highest point in its trajectory and it is fired at an initial velocity of 60 m / s at an angle of 70 degrees from the horizontal.

How would you find out what height H it explodes at? And what time is it from the start when it explodes?

This is one of many problems affecting the maximum height of a projectile and the trick to solving these problems is that at maximum height the y-Component of the speed is 0 m / s for a moment. By plugging in that value for vy By choosing the most appropriate kinematic equation, you can easily solve this and similar problems.

Looking at the kinematic equations first, this one pops out (with indices added to show that we are working in the vertical direction):

v_y ^ 2 = v_ {0y} ^ 2 + 2a_ys_y

This equation is ideal because you already know the acceleration (ay = -G), the initial velocity, and the initial angle (so that you can calculate the vertical component vy0). As we are looking for the value of sy (i.e. the height H) when vy = 0, we can replace the final vertical velocity component with zero and rearrange it sy:

0 = v_ {0y} ^ 2 + 2a_ys_y −2a_ys_y = v_ {0y} ^ 2 s_y = frac {−v_ {0y} ^ 2} {2a_y}

It makes sense to name the upward direction yand since the acceleration of gravity G directed downwards (i.e. in the -y Direction) we can change ay to the -G. Finally call sy the height H, we can write:

h = frac {v_ {0y} ^ 2} {2g}

So, to solve the problem, all you need to do is calculate the vertical component of the initial speed, which you can do using the trigonometric approach from the previous section. With the information from the question (60 m / s and 70 degrees to the horizontal start) we get:

begin {align} v_ {0y} & = 60; {m / s} x sin (70) & = 56.38; {m / s} end {align}

Now you can search for the maximum height:

begin {align} h & = frac {v_ {0y} ^ 2} {2g} & = frac {(56.38; {m / s}) ^ 2} {2 × 9.8; {m / s} ^ 2} & = 162.19 {m} end {align}

The fireworks explode at a height of around 162 meters.

### Continuation of the example: flight time and distance traveled

After solving the basics of the projectile movement problem which is based solely on vertical movement, the rest of the problem can be solved easily. First, the time from the start that the fuse blows can be determined using one of the other constant acceleration equations. If you look at the options you will see the following expression:

s_y = bigg (frac {v_y + v_ {0y}} {2} bigg) t

has the time t, what do you want to know; the displacement you know for the maximum point of flight; the initial vertical speed; and the speed at the time of maximum altitude (which we know is zero). Based on this, the equation can be rearranged to give an expression for the time of flight:

s_y = bigg (frac {v_ {0y}} {2} bigg) t t = frac {2s_y} {v_ {0y}}

So insert values ​​and search t gives:

begin {align} t & = frac {2 × 162.19; {m}} {56.38; {m / s}} & = 5.75; {s} end {align}

So the fireworks will explode 5.75 seconds after launch.

Finally, you can easily find the horizontal distance traveled using the first equation, which is (in the horizontal direction):

v_x = v_ {0x} + a_xt

However, it should be noted that there is no acceleration in the X-direction, that's easy:

v_x = v_ {0x}

That means the speed in the X The direction is the same throughout the fireworks trip. Given that v = d/t, Where d is the distance traveled, this is easy to see d = vtand so in this case (with sX = d):

s_x = v_ {0x} t

So you can replace v0x Using the trigonometric expression from earlier, enter the values ​​and solve:

begin {align} s_x & = v_0 cos (θ) t & = 60; {m / s} × cos (70) × 5.75; {s} & = 118; {m} end {aligned}

So it will travel about 118 m before the explosion.

### Additional projectile movement problem: The dud fireworks

For an additional problem, imagine that the fireworks from the previous example (initial speed of 60 m / s, started at 70 degrees to the horizontal) did not explode at the height of its parabola and instead landed on the ground without exploding. In this case, can you calculate the total flight time? How far is it from the launch site in the horizontal direction, will it land, or in other words what is it? Range of the projectile?

This problem works in principle the same way, with the vertical components of speed and displacement being the most important factors to consider in determining flight time and from which you can determine range. Instead of working through the solution in detail, you can solve it yourself using the previous example.

There are formulas for the range of a projectile that you can look up or derive from the constant acceleration equations. However, this is not really necessary as you already know the maximum height of the projectile and from that point on you will only be in free fall under the action of gravity.

That said, you can determine the time it takes for the fireworks to fall back to the ground and then add that to the flight time to the maximum altitude to determine the total flight time. From this point on, the constant speed in the horizontal direction is used in addition to the flight time to determine the range.

Show that the flight time is 11.5 seconds and the range is 236 m. Note that you need to calculate the vertical component of velocity at the point where it hits the ground as an intermediate step.