Projectile motion

Parabolic water trajectory

Initial velocity of parabolic throwing

Components of initial velocity of parabolic throwing

Projectile motion is a form of motion where a particle (called a projectile) is thrown obliquely near the earth's surface, & it moves along a curved path under the action of gravity. The path followed by a projectile motion is called its trajectory. Projectile motion only occurs when there is one force applied at the beginning of the trajectory after which there is no interference apart

The initial velocity

If the projectile is launched with an initial velocity v₀, 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, ϴ 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 by

$V_{0}={\sqrt {{R^{2}g} \over {R\sin 2\theta +2h\cos ^{2}\theta }}}$.

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 no acceleration in the horizontal direction velocity in horizontal direction is constant which is equal to ucosα. The vertical motion of the projectile is the motion of a particle during its free fall. Here the acceleration is constant, equal to g. The components of the acceleration:

$a_{x}=0$,

$a_{y}=g$.

Velocity

The horizontal component of the velocity remains unchanged throughout the motion. The vertical component of the velocity increases linearly, because the acceleration is constant. At any time t, the components of the velocity:

$v_{x}=v_{0}\cos(\theta )$,

$v_{y}=v_{0}\sin(\theta )-gt$.

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

Consider the equations,

$x=v_{0}t\cos(\theta )$,

$y=v_{0}t\sin(\theta )-{\frac {1}{2}}gt^{2}$.

If we eliminate t between these two equations we will obtain the following:

$y=\tan(\theta )\cdot x-{\frac {g}{2v_{0}^{2}\cos ^{2}\theta }}\cdot x^{2}$,

This equation is the equation of a parabola. Since g, α, and v₀ 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.

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:

$t_{h}={v_{0}\sin(\theta ) \over g}$.

From the vertical displacement the maximum height of projectile:

$h=v_{0}t_{h}\sin(\theta )-{\frac {1}{2}}gt_{h}^{2}$

$h={v_{0}^{2}\sin ^{2}(\theta ) \over {2g}}$ .

Additional Equation

For the relation between the distance traveled, the maximum height and angle of launch, the equation below has been developed.

$h={d\tan(\theta ) \over {4}}$ Thus we can easily define the path of the projectile.

The maximum distance of projectile

The maximum distance 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).

$0=v_{0}t_{d}\sin(\alpha )-{\frac {1}{2}}gt_{d}^{2}$.

Time to reach ground:

$t_{d}={2v_{0}\sin(\alpha ) \over g}$.

From the horizontal displacement the maximum distance of projectile:

$d=v_{0}t_{d}\cos(\alpha )$,

so

$d={\frac {v_{0}^{2}}{g}}\sin(2\alpha )$.

Note that d has its maximum value when

$\sin 2\alpha =1$,

which necessarily corresponds to

$2\alpha =90^{\circ }$,

or

$\alpha =45^{\circ }$.

Other useful equation

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

$v_{y}^{2}=(v_{0}\sin \alpha )^{2}-2gy$.

References

• Budó Ágoston: Kísérleti fizika I.,Budapest, Tankönyvkiadó, 1986. ISBN 963 17 8772 9 Template:Hu
• Ifj. Zátonyi Sándor: Fizika 9.,Budapest, Nemzeti Tankönyvkiadó, 2009. ISBN 978-963-19-6082-2 Template:Hu
• Hack Frigyes: Négyjegyű függvénytáblázatok, összefüggések és adatok, Budapest, Nemzeti Tankönyvkiadó, 2004. ISBN 963-19-3506-X Template:Hu

External links

• Open Source Physics computer model
• A Java simulation of projectile motion, including first-order air resistance

NOTE: Since the value of g is not specific the body with high velocity over g limit cannot be measured using the concept of the projectile motion.