projectile motion

Consider the motion of a particle which is projected in a direction making an angle $ \alpha$ with the horizon. When we neglect drag, the only force which acts upon the particle is its weight, $ m{\bf g}$ (Fig. 66).

\includegraphics[scale=.8]{Fig66.eps}

Taking the plane of motion to be the xy-plane, and applying Newton's laws of motion gives us the equations

x-axis

$\displaystyle m \frac{d \ddot{x}}{dt} = 0$

$\displaystyle \frac{d \ddot{x}}{dt} = 0$ (1)

y-axis

$\displaystyle m \frac{d \ddot{y}}{dt} = -mg$

$\displaystyle \frac{d \ddot{y}}{dt} = -g$ (2)

where $ \frac{d \ddot{x}}{dt}$ and $ \frac{d \ddot{y}}{dt}$ are the components of the acceleration along the x and y axes. Integrating equations (1) and (2) we get

$\displaystyle \dot{x} = c_1 $

$\displaystyle \dot{y} = -gt + c_2 $

Therefore the component of the velocity along the x-axis remains constant, while the component along the y-axis changes uniformly. Let $ v_0$ be the initial velocity of the projection, then when $ t = 0$ , $ \dot{x}_0 = v_0 \cos \alpha$ and $ \dot{y}_0 = v_0 \sin \alpha$ . Making these substitutions in the last two equations we obtain

$\displaystyle c_1 = v_0 \cos \alpha $

$\displaystyle c_2 = v_0 \sin \alpha $

Therefore

$\displaystyle \dot{x} = v_0 \cos \alpha$ (3)

$\displaystyle \dot{y} = v_0 \sin \alpha - gt$ (4)

Then the total velocity at any instant is

$\displaystyle v = \sqrt{\dot{x}^2 + \dot{y}^2} $

$\displaystyle v = \sqrt{v_0^2 - 2v_0 g t \sin \alpha + g^2 t^2} $

and makes an angle $ \theta$ with the horizon defined by

$\displaystyle \tan \theta = \frac{ \dot{y} }{ \dot{x} } = \frac{ v_0 \cos \alpha}{ v_0 \sin \alpha - gt} $

Integrating equations (3) and (4) we obtain

$\displaystyle x = v_0 t \cos \alpha + c_3 $

$\displaystyle y = v_0 t \sin \alpha - \frac{1}{2} g t^2 + c_4 $

But when $ t = 0$ , $ x = y = 0 $ , therefore $ c_3 = c_4 = 0$ , and consequently

$\displaystyle x = v_0 t \cos \alpha$ (5)

$\displaystyle y = v_0 t \sin \alpha - \frac{1}{2} g t^2$ (6)

It is interesting to note that the motions in the two directions are independent. The gravitational acceleration does not affect the constant velocity along the x-axis, while the motion along the y-axis is the same as if the body were dropped vertcally with an initial velocity $ v_0 \sin \alpha $ .

bf The Path - The equation of the path may be obtained by eliminating $ t$ between equations (7) and (8). This gives

$\displaystyle y = x \tan \alpha - \frac{g}{2 v_0^2 \cos^2 \alpha} x^2$ (7)

which is the equation of a parabola.

bf The Time of Flight - When the projectile strikes the ground its y-coordinate is zero. Therefore substituting zero for $ y$ in equation (8) we get for the time of flight

$\displaystyle T = \frac{2 v_0 \sin \alpha}{g}$ (8)

The Range - The range, or the total horizontal distance covered by the projectile, is found by replacing $ t$ in equation (7) by the value of $ T$ in equation (10), or by letting $ y = 0$ in equation (9). By either method we obtain

$\displaystyle R = \frac{2 v_0^2 \sin \alpha \cos \alpha}{g} = \frac{v_0^2 \sin 2 \alpha}{g}$ (9)

Note that a basic trigonometric identity was used to simpilfy the above equation.

Since $ v_0$ and $ g$ are constants the value of $ R$ depends upon $ \alpha$ . It is evident from equation (11) that $ R$ is maximum when $ \sin 2 \alpha = 1$ , or when $ \alpha = \frac{\pi}{4}$ . The maximum range is, therefore,

$\displaystyle R_{max} = \frac{v_0^2}{g}$ (10)

In actual practice the angle of elevation which gives the maximum range is smaller on account of the resistance of the air.

The Highest Point - At the highest point $ \dot{y}=0$ . Therefore substituting this value of $ \dot{y}$ in equation (4) we obtain $ \frac{v_0 \sin \alpha}{g}$ or $ \frac{1}{2}T$ for the time taken to reach the highest point. Subsituting this value of the time in equation (8) we get for the maximum elevation

$\displaystyle H = \frac{v_0^2 \sin^2 \alpha}{2 g}$ (11)

The Range for a Sloping Ground - Let $ \beta$ be the angle which the ground makes with the horizon. Then the range is the distance $ OP$ , Fig. 67, where $ P$ is the point where the projectile strikes the sloping ground. The equation of the line $ OP$ is

$\displaystyle y = x \tan \beta$ (12)

\includegraphics[scale=.8]{Fig67.eps}

Eliminating $ y$ between equations (14) and (9) we obtain the x-coordinate of the point,

$\displaystyle x_p = \frac{ 2 v_0^2 \cos^2 \alpha \left ( \tan \alpha - \tan \beta \right ) }{g} $

But $ x_p = R' \cos \beta$ , where $ R' = OP$ .

Therefore

$\displaystyle R' = \frac{2 v_0^2 \cos \alpha}{g \cos^2 \beta} \sin \left ( \alpha - \beta \right ) $

$\displaystyle R' = \frac{ v_0^2}{g} \frac{ \sin \left ( 2 \alpha - \beta \right ) - \sin \beta }{ \cos^2 \beta }$ (13)

Thus for a given value of $ \beta$ , $ R'$ is maximum when $ \sin \left ( 2 \alpha - \beta \right ) = 1$ , that is, when $ \alpha = \frac{\pi}{4} + \frac{\beta}{2}$ .

$\displaystyle R_{max}' = \frac{ v_0^2 1 - \sin \beta}{ g \cos^2 \beta} = \frac{v_0^2}{g \left ( 1 + \sin \beta \right ) }$ (14)

When $ \beta = 0$ equations (15) and (16) reduce to equations (12) and (13), as they should.



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As of this snapshot date, this entry was owned by bloftin.