Projectile motion

The initial velocity

Let the projectile be launched with an initial velocity v ( 0 ) ≡ v 0 , which can be expressed as the sum of horizontal and vertical components as follows:

v 0 = v 0 x i + v 0 y j .

The components v 0 x and v 0 y can be found if the initial launch angle, θ , is known:

v 0 x = v 0 cos ⁡ θ , v 0 y = v 0 sin ⁡ θ .

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. This is the principle of compound motion established by Galileo in 1638.

Acceleration

Since there is only acceleration in the vertical direction, the velocity in the horizontal direction is constant, being equal to v 0 cos ⁡ θ . 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 . 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 downward 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 ⁡ ( θ ) , v y = v 0 sin ⁡ ( θ ) − g t .

The magnitude of the velocity (under the Pythagorean theorem, also known as the triangle law):

v = v x 2 + v y 2   .

Displacement

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

x = v 0 t cos ⁡ ( θ ) , y = v 0 t sin ⁡ ( θ ) − 1 2 g t 2 .

The magnitude of the displacement is:

Δ r = x 2 + y 2   .

Consider the equations,

x = v 0 t cos ⁡ ( θ ) , y = v 0 t sin ⁡ ( θ ) − 1 2 g t 2 .

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

y = tan ⁡ ( θ ) ⋅ x − g 2 v 0 2 cos 2 ⁡ θ ⋅ x 2 .

Since g , θ , and v 0 are constants, the above equation is of the form

y = a x + b x 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.

If the projectile's position (x,y) and launch angle (θ or α) are known, the initial velocity can be found solving for v 0 in the aforementioned parabolic equation:

v 0 = x 2 g x sin ⁡ 2 θ − 2 y cos 2 ⁡ θ .

Time of flight or total time of the whole journey

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

y = v 0 t sin ⁡ ( θ ) − 1 2 g t 2

After the flight, the projectile returns to the horizontal axis (x-axis), so y=0

0 = v 0 t sin ⁡ ( θ ) − 1 2 g t 2 v 0 t sin ⁡ ( θ ) = 1 2 g t 2 v 0 sin ⁡ ( θ ) = 1 2 g t t = 2 v 0 sin ⁡ ( θ ) g

Note that we have neglected air resistance on the projectile.

Maximum height of projectile

The greatest height that the object will reach is known as the peak of the object's motion. The increase in height will last until v y = 0 , that is,

0 = v 0 sin ⁡ ( θ ) − g t h .

Time to reach the maximum height(h):

t h = v 0 sin ⁡ ( θ ) g .

From the vertical displacement of the maximum height of projectile:

h = v 0 t h sin ⁡ ( θ ) − 1 2 g t h 2 h = v 0 2 sin 2 ⁡ ( θ ) 2 g .

Relation between horizontal range and maximum height

The relation between the range R on the horizontal plane and the maximum height h reached at t d 2 is:

h = R tan ⁡ θ 4

Proof

h = v 0 2 sin 2 ⁡ θ 2 g

R = v 0 2 sin ⁡ 2 θ g h R = v 0 2 sin 2 ⁡ θ 2 g × g v 0 2 sin ⁡ 2 θ h R = sin 2 ⁡ θ 4 sin ⁡ θ cos ⁡ θ

h = R tan ⁡ θ 4 .

Maximum distance of projectile

It is important to note that the range and the maximum height of the projectile does not depend upon its mass. Hence range and maximum height are equal for all bodies that are thrown with the same velocity and direction.

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

0 = v 0 t d sin ⁡ ( θ ) − 1 2 g t d 2 .

Time to reach ground:

t d = 2 v 0 sin ⁡ ( θ ) g .

From the horizontal displacement the maximum distance of projectile:

d = v 0 t d cos ⁡ ( θ ) ,

so

d = v 0 2 g sin ⁡ ( 2 θ ) .

Note that d has its maximum value when

sin ⁡ 2 θ = 1 ,

which necessarily corresponds to

2 θ = 90 ∘ ,

or

θ = 45 ∘ .

Application of the work energy theorem

According to the work-energy theorem the vertical component of velocity is:

v y 2 = ( v 0 sin ⁡ θ ) 2 − 2 g y .