Projectile motion

Parabolic water trajectory

Initial velocity of parabolic throwing

Components of initial velocity of parabolic throwing

Projectile motion is a form of motion in which an object or particle (called a projectile) is thrown near the earth's surface, and it moves along a curved path under the action of gravity only. The only force of significance that acts on the object is gravity, which acts downward to cause a downward acceleration. There are no horizontal forces needed to maintain the horizontal motion – consistent with the concept of inertia.

The initial velocity

If the projectile is launched with an initial velocity $\mathbf{v}_0$ , then it can be written as

\mathbf{v}_0 = v_{0x}\mathbf{i} + v_{0y}\mathbf{j}

The components $v_{0x}$ and $v_{0y}$ can be found if the angle, $\theta$ is known:

v_{0x} = v_0\cos\theta

v_{0y} = v_0\sin\theta

If the projectile's range, launch angle, and drop height are known, launch velocity can be found using Newton's formula

V_0 = \sqrt{{R^2 g} \over {R \sin 2\theta + 2h \cos^2\theta}}

The launch angle is usually expressed by the symbol theta, but often the symbol alpha is used.

Kinematic quantities of projectile motion

In projectile motion, the horizontal motion and the vertical motion are independent of each other; that is, neither motion affects the other.

Acceleration

Since there is only acceleration in the vertical direction, the velocity in the horizontal direction is constant, being equal to $\mathbf{v}_0 \cos(\theta)$ . The vertical motion of the projectile is the motion of a particle during its free fall. Here the acceleration is constant, being equal to $g$ .^[1] The components of the acceleration are:

a_x = 0

a_y =- g

Velocity

The horizontal component of the velocity of the object remains unchanged throughout the motion. The vertical component of the velocity increases linearly, because the acceleration due to gravity is constant. The accelerations in the $x$ and $y$ directions can be integrated to solve for the components of velocity at any time $t$ , as follows:

v_x=v_0 \cos(\theta)

v_y=v_0 \sin(\theta) - gt

The magnitude of the velocity (under the Pythagorean theorem):

v=\sqrt{v_x^2 + v_y^2 \ }

Displacement

Displacement and coordinates of parabolic throwing

At any time $t$ , the projectile's horizontal and vertical displacement:

x = v_0 t \cos(\theta)

y = v_0 t \sin(\theta) - \frac{1}{2}gt^2

The magnitude of the displacement:

\Delta r=\sqrt{x^2 + y^2 \ }

Parabolic trajectory

Main article: Trajectory of a projectile

Consider the equations,

x = v_0 t \cos(\theta)

y = v_0 t \sin(\theta) - \frac{1}{2}gt^2

If t is eliminated between these two equations the following equation is obtained:

y=\tan(\theta) \cdot x-\frac{g}{2v^2_{0}\cos^2 \theta} \cdot x^2

This equation is the equation of a parabola. Since $g$ , $\theta$ , and $\mathbf{v}_0$ are constants, the above equation is of the form

y=ax+bx^2

in which $a$ and $b$ are constants. This is the equation of a parabola, so the path is parabolic. The axis of the parabola is vertical.

Time of Flight

The total time for which the projectile remains in the air is called the time of flight.

y = v_0 t \sin(\theta) - \frac{1}{2}gt^2

After the flight, the projectile returns to the horizontal axis, so y=0

0 = v_0 t \sin(\theta) - \frac{1}{2}gt^2

v_0 t \sin(\theta) = \frac{1}{2}gt^2

v_0 \sin(\theta) = \frac{1}{2}gt

t = \frac{2 v_0 \sin(\theta)}{g}

Note that we have neglected air resistance on the projectile.

The maximum height of projectile

Maximum height of projectile

The highest height which the object will reach is known as the peak of the object's motion. The increase of the height will last, until $v_y=0$ , that is,

0=v_0 \sin(\theta) - gt_h

Time to reach the maximum height:

t_h = \frac{v_0 \sin(\theta)}{g}

From the vertical displacement of the maximum height of projectile:

h = v_0 t_h \sin(\theta) - \frac{1}{2} gt^2_h

h = \frac{v_0^2 \sin^2(\theta)}{2g}

Relation during horizontal range and maximum height

The relation between the range $R$ on the horizontal plane and the maximum height $h$ reached at $\frac{t_d}{2}$ is:

h = \frac{R\tan\theta}{4}

The maximum distance of projectile

Main article: Range of a projectile

The maximum distance of projectile

It is important to note that the Range and the Maximum height of the Projectile does not depend upon mass of the trajected body. Hence Range and Maximum height are equal for all those bodies which are thrown by same velocity and direction. Air resistance does not affect displacement of projectile.

The horizontal range d of the projectile is the horizontal distance the projectile has travelled when it returns to its initial height (y = 0).

0 = v_0 t_d \sin(\theta) - \frac{1}{2}gt_d^2

reach ground:

t_d = \frac{2v_0 \sin(\theta)}{g}

From the horizontal displacement the maximum distance of projectile:

d = v_0 t_d \cos(\theta)

so^[2]

d = \frac{v_0^2}{g}\sin(2\theta)

Note that $d$ has its maximum value when

\sin 2\theta=1

which necessarily corresponds to

2\theta=90^\circ

\theta=45^\circ

Application of the work energy theorem

According to the work-energy theorem the vertical component of velocity are listed below:

v_y^2 = (v_0 \sin \theta)^2-2gy