ProjectileMotion

An open source physics library

create new user


Username:
Password:

forget your password?

Main Menu

Sections

Encyclopedia

Papers

Books

lectures

Talkback

Polls

Forums

Bug Reports

Feedback

Downloads

Snapshots

Information

Legalese

About

projectile motion

(Definition)

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.

"projectile motion" is owned by bloftin.