Air Resistance Formula
We have seen the planes moving in sky what decreases its speed? This is what Air resistance is!
[pic]
Air resistance is the resistance against the air, which decreases the speed of the moving object. Its formula is given as
[pic]
Where c is air constant, v is the object's velocity.
Air Resistance Formula is used to find the air resistance, air constant and velocity of body if some of these quantities are known. This formula has wide applications in aeronautics.

Air resistance Problems

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Below are given some problems based on air resistance which may be helpful for you.

Solved Examples

Question 1: A plane moving with velocity of 50 ms-1 has a force constant of 0.05. Calculate its air resistance.
Solution:

Given: Velocity of air v = 50 ms-1, Force constant c = 0.05
The force constant is given by F = cv2 = 0.05 × 2500 = 125 N.

Question 2: An object is traveling at 20 ms-1 experiences a force of 50 N. Calculate the force constant.
Solution:

Given: Velocity of air v = 20 ms-1, Force F = 50 N,
The force constant is given by c = F/v² = 50/20² = 0.125

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Projectile Motion

The motion of object in two dimensions is explained by two main principles that are kinematic principles and Newton's laws of motion. The motion in two dimensions is called the motion of projectiles. The projectile object is one on which is moved under the force of gravity. Some examples of projectiles are like an object is dropped from its rest condition on which there is no effect of air resistance. An object is thrown from vertical under negligible effect of air resistance an object and if it is thrown in up direction at a horizontal angle is considered in a projectile motion.

So, we can say that a projectile is any object which is projected continues in motion under its inertia and force of gravity. Let’s discuss the definition of projectile, its range equations, horizontal motion equation and reason for this motion, motion with air resistance, and problem based on it.

[pic]

Here a welder cuts holes through a heavy metal construction beam with a hot torch. The sparks generated in the process follow parabolic path.

Projectile Motion Definition

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Projectile motion is an example of curved motion with constant acceleration. It is two dimensional motion of a particle thrown obliquely into the air.

Consider the motion and path followed by the ball when it moves in the curved path. We will make two assumptions here:

a) First assumption is that the free fall acceleration (g) remains constant and does not change its value during the motion of the ball.

b) Resistance offered by the ball is negligible.

If we consider the motion and the assumptions stated above, we will find that :

1. The path of the projectile (ball here) is always a parabola. 2. The path followed by the projectile is termed as the "trajectory of the projectile". 3. Projectile feels only one force while in motion, which is the force of gravity.

Projectile Motion is Caused by?

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Projectile motion is caused by the gravitational force of earth.

There could be different projectile motions,

• An object thrown from a hill to the downward direction, considering that initially the object was at rest, is a projectile motion. We are not considering the effect of the air resistance here. Object will fall down towards the center of the earth due to the force of the gravity. • An object thrown from the ground towards the sky or in the upward direction, follows the projectile motion. Initially a force is applied to the object and its initial velocity is not zero. We are not considering the effect of air resistance here. • An object, thrown towards the sky but by making some angle with horizontal surface, follows the curved path and also the projectile motion. Here also we are neglecting the effect of air resistance.

[pic]

Figure 1: General Trajectory of the Projectile

Projectile Motion Equation

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Projectile motion is a two dimensional concept and it follows the two dimensional kinematics. A projectile has both the horizontal and the vertical components of motion.

Projectile motion can be stated as the:

y = 12 (at2) + v0 t + y0
Where,
y = height t = time a = acceleration of the projectile because of gravity
V0 = Initial velocity of the projectile
Y0 = Initial height of the projectile

Horizontal Component of the Velocity : Whenever the projectile is thrown or follows the trajectory, the horizontal component of the velocity does not changes and the displacements covered by the horizontal components of the velocity are uniform. In other words, final horizontal velocity component is equal to the initial velocity component.

Now the point here to note is that when the projectile follows the trajectory, gravity force does not affect or does not make any change in the horizontal velocity component of the velocity.

Vertical Component of the Velocity : Vertical component of the velocity does not remain constant during the projectile motion. Gravity force acts on it and changes the vertical component of the velocity of the projectile. The displacements covered by the vertical component of the velocity are not uniform.

For the vertical component of the velocity during the projectile motion, change in both the magnitude and direction takes place. If the projectile is moving in the upward direction, then the vertical component of the velocity is in the upward direction and decrease in its magnitude takes place.

On the other hand, when the projectile moves in the downward direction, the direction of the vertical components of the velocity is in the downward direction and increase in the magnitude takes place.

Range Equation for Projectile Motion

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|Equations involving Vertical |Equations involving Horizontal |Explanation of Symbols used |
|Motion |Motion | |
| | | |
|V(iy) = Vi sinθ |V(ix) = Vi cosθ |Vi = Magnitude of Initial Velocity |
| | |V(iy) = Y component of Initial Velocity |
| | |V(ix) = X component of Initial Velocity |
|V(fy) = V(iy) + ay t |V(fx) = V(ix) |Vfy = Y component of Velocity at time 't' |
| | |Vfx = X component of Velocity at time 't' (note that the X component remains constant) |
| | |ay = acceleration in vertical direction, which in the case of projectile motion would be|
| | |-9.8 m/s2 |
|Yf - Yi = V(iy) t + 12 ay t2 |Xf - Xi = V(ix) t |Yi = Initial Y co-ordinate of Projectile |
| | |Yf = Y co-ordinate of Projectile at time 't' |
| | |Xi = Initial X co-ordinate of Projectile |
| | |Xf = X co-ordinate of Projectile at time 't' |
| | |Symbols already described above |
|(Vfy)2 = (V(iy))2 + 2ay(Yf - Yi) | | |
| | |Symbols already described above |
|Yf - Yi = t.(V(iy)+V(fy))2 | | |

Maximum Projectile Range : Expression

Now, lets look at the expression for projectile range using the above formula, Let the projectile start at (0, Yi) co-ordinates with a speed of Vi = v, and angle θ with the horizontal surface. After some time t, it strikes the ground at a distance of Xf. The value of Xf gives the range of the projectile

The figure given below aids the visualization of the motion :

[pic]

In this figure, the range of the projectile is given by the formula,

d = Xf = ((Vcosθ)g) (Vsinθ+((Vsinθ)2+2gYi)−−−−−−−−−−−−−−√)

Using the above equation one can make a graph of `theta` versus `d` for different `theta`, and see where the value of `d` maximizes. This will be the value of maximum projectile range. Moreover, this equation reduces to a very simple form when the projectile starts form ground level, that is when Yi = 0.

The equation then becomes :

d = Xf = (2V2cosθsinθ)g

Which can then be simplified to the following expression :

d = Xf = (V2sin(2θ))g

Using the above equation we can very easily find the expression for maximum projectile range in this simple situation. We know that the maximum value of sin 2θ is 1.

Therefore, the maximum range of the projectile is

d = Xf = V2g

Also, the value of 2θ for which sin 2θ = 1 is 90∘. Therefore, the value of θ = 90/2 = 45∘

Horizontal Projectile Motion

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This is a type of Projectile motion in which projectile does not follow path in the upward direction or it does not have upward trajectory and the initial velocity of the projectile is also zero. This type of projectile motion is called horizontal projectile motion. This motion generally occurs when the projectile is shot straight without forming any angle with the horizontal surface and the projectile falls downward until it hits the ground.

Exemplary Horizontal Projectile motion is shown in the figure below.

As shown in the figure below, the initial component of the vertical components of the velocity is zero. Horizontal velocity component of the projectile remains constant as the gravity does not affect it. Direction of the vertical component of the velocity is in downward direction during the trajectory. The magnitude of the vertical component of the velocity increases as the projectile moves downward, the force of gravity acts on it, results in acceleration of the projectile.

[pic]

The figure above illustrates a body thrown horizontally from a point O with a velocity [pic] The point O is at a certain height above the ground. Let x and y be the horizontal and vertical distances covered by the projectile, respectively, in time t. Therefore, at time t, the projectile is at p.
In order to calculate x, let us consider the horizontal motion, which is uniform motion. This is because the only force acting on the projectile is the force of gravity. This force acts vertically downwards and hence the horizontal component in zero. Therefore, the equations of motion of the projectile for the horizontal direction is just the equation of uniform motion in a straight line.
∴ x = vt ------------------ (i)
In order to calculate y, the vertical motion of the projectile is considered. Since the vertical motion is controlled by the force of gravity, it is an accelerated motion. The initial velocity, vy (0), in the vertically downward direction is zero. Since the Y-axis in the figure above is taken downwards, the downward direction is regarded as the positive direction. So, the acceleration of the projectile is + g.
∴ from the equation y(t) = Vy(0)t + 12 ay t2
We have y(t) = 12 gt2 -------------(2)
Here vy (0) is taken as zero because both distance and time are being measured from the origin O.
From equation (1) t = xv
Substituting for t from the above equation in equation (2) we have, y(t) = 12 g(xv)2 = g2v2 x2
∴ y = kx2 Where k = g2v2 .................(3) is a constant for a projectile projected upwards with a definite velocity v and at a place with a definite value of 'g'.
Equation (3) is a second-degree equation in x, a first-degree equation in y and is the equation of a parabola. Therefore, a body thrown horizontally from a certain height above the ground follows a parabolic trajectory till it hits the ground.
Resultant Velocity of a Horizontal Projectile:
In this section, let us calculate the resultant velocity of the projectile V⃗ , at any point p on the trajectory, in an interval of time t. Vx and Vy are the horizontal and vertical components of V⃗ as illustrated in the figure below.

[pic]

Since, the horizontal motion of the projectile is uniform, V⃗ x = V⃗
However, the motion in the vertical direction is an acceleration one.
∴ Vy(t) = Vy (0) + ay t
Since O is considered to be the origin, Vy (0) = 0
∴ Vv (t) = gt
∴ The magnitude of the resultant velocity V⃗ is given by,
|V⃗ | = V = V2x+V2y−−−−−−−√
∴ V = V2+g2t2−−−−−−−−√
The direction is given by tanβ = VyVx = gtV
∴ β = tan-1gtV

Projectile Motion Equations Max Height

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Maximum height covered by the projectile is given by:

hmax = u2v19.6

where,

uv is the initial vertical velocity hmax = maximum height covered by the projectile

Projectile Motion with Air Resistance

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So far, in the study of projectile motion, we have not considered the effect of air resistance on the projectile. The only force considered is the gravity force.

In case of solving the projectile motion equations, the drag provided by the air resistance is also considered.

Projectile Motion Simulator

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Projectile motion simulators are the type of softwares which are used to study the projectile motion of the different objects depending on the various conditions of :

• Initial velocity given to the projectile or initial force applied to the projectile. • Studying the trajectory followed by the projectile considering different sources of resistance, e.g., air resistance. • To study the projectile behavior of various objects. • To analyze and find how much force should be applied to the projectile for a particular desired trajectory.

Projectile Motion Problems

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Problems on projectile motion are given below:

Solved Examples

Question 1: Find the maximum range of a projectile launched at the speed of 25 m/s. Also find the range when the launch angle is 30 degrees?
Solution:

The maximum range is d = (V2g) = 2529.8 = 63.8 m

Projectile Range when launch angle is
30 = d = (V2sin(2θ))g = (252sin(2×30))9.8 = 55.2 m

Question 2: Find the launch angle of a projectile launched with a speed of 40 m/s, where the projectile range is 60 m?
Solution:

Projectile range d = (V2sin(2θ))g
⇒ 60 = (402sin(2θ))9.8
⇒ sin(2θ) = 0.3675
⇒ 2θ = sin-1(0.3675) = 21.56 or 180 - 21.56
⇒ θ = 10.8∘ or (90∘ - 10.8∘) = 79.2∘
[θ] , there is another launch angle [90-θ] for which the projectile range is same.

Question 3: If Football is kicked by the player at the velocity of 20.0 m/s with the initial angle of 53 degrees, then what is the range covered by the football?
Solution:

According to the projectile motion equation the range covered by the projectile would be
= Initial velocity × cos(53)
= 20 × cos(53)
= 12.04 m/sec

Air resistance calculator

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