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 v0x and v0y can be found if the angle, ϴ is known:
v_{0x} = vgh_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.[1] 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^2_{0}\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^2_h 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}}
References
Budó Ágoston:, összefüggések és adatok, Budapest, Nemzeti Tankönyvkiadó, 2004. ISBN 963-19-3506-X (Hungarian)
Notes
^ The g is the acceleration due to gravity. (9.81 m/s2 near the surface of the Earth).
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.
A projectile is any object upon which the only force is gravity,
Projectiles travel with a parabolic trajectory due to the influence of gravity,
There are no horizontal forces acting upon projectiles and thus no horizontal acceleration,
The horizontal velocity of a projectile is constant (a never changing in value),
There is a vertical acceleration caused by gravity; its value is 9.8 m/s/s, down,
The vertical velocity of a projectile changes by 9.8 m/s each second,
The horizontal motion of a projectile is independent of its vertical motion.
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