Controlo de Variáveis

Como podemos controlar uma determinada variável?

Intensidade de v0 / ms-1 Ângulo / º Alcance (x) / m altura max. ( h

max) / m

10 10 ? ?

20 10 ? ?

30 10 ? ?

20 20 ? ?

20 45 ? ?

20 60 ? ?

Estudar a variável v0, mantendo a

variável ângulo

controlada

Estudar a variável ângulo, mantendo

a variável v0 controlada

S E C T I O N 4 . 3 • Projectile Motion 87

x(m)

50

100

150

y(m)

75°

60°

45°

30°

15°

vi = 50 m/s

50 100 150 200 250

Active Figure 4.11 A projectile launched from the origin with an initial speed of50 m/s at various angles of projection. Note that complementary values of !i result inthe same value of R (range of the projectile).

P R O B L E M - S O LV I N G H I N T S

Projectile MotionWe suggest that you use the following approach to solving projectile motionproblems:

• Select a coordinate system and resolve the initial velocity vector into x and ycomponents.

• Follow the techniques for solving constant-velocity problems to analyze the hori-zontal motion. Follow the techniques for solving constant-acceleration problemsto analyze the vertical motion. The x and y motions share the same time t.

Example 4.3 The Long Jump

A long-jumper (Fig. 4.12) leaves the ground at an angle of20.0° above the horizontal and at a speed of 11.0 m/s.

(A) How far does he jump in the horizontal direction? (As-sume his motion is equivalent to that of a particle.)

Solution We conceptualize the motion of the long-jumperas equivalent to that of a simple projectile such as the ballin Example 4.2, and categorize this problem as a projectilemotion problem. Because the initial speed and launch an-gle are given, and because the final height is the same asthe initial height, we further categorize this problem assatisfying the conditions for which Equations 4.13 and4.14 can be used. This is the most direct way to analyze thisproblem, although the general methods that we have beendescribing will always give the correct answer. We will takethe general approach and use components. Figure 4.10

Quick Quiz 4.6 Rank the launch angles for the five paths in Figure 4.11 withrespect to time of flight, from the shortest time of flight to the longest.

provides a graphical representation of the flight of thelong-jumper. As before, we set our origin of coordinates atthe takeoff point and label the peak as ! and the landingpoint as ". The horizontal motion is described by Equa-tion 4.11:

The value of x B can be found if the time of landing tB isknown. We can find tB by remembering that ay " # g and byusing the y part of Equation 4.8a. We also note that at the topof the jump the vertical component of velocity vyA is zero:

tA " 0.384 s

0 " (11.0 m/s) sin 20.0$ # (9.80 m/s2)tA

vyf " vyA " vi sin !i # gtA

xf " x B " (vi cos !i)t B " (11.0 m/s)(cos 20.0$)t B

At the Active Figures link athttp://www.pse6.com, you canvary the projection angle to ob-serve the effect on the trajectoryand measure the flight time.

Pretende-se estudar o comportamento de que grandezas?

Qual a grandeza que é controlada?

Qual a grandeza que está a variar?

Movimento de Projéteis

Movimento em 2 Dimensões

84 C H A P T E R 4 • Motion in Two Dimensions

xvxi

vyi v

vxi

θvy v

gvxivy = 0

vxi

vyv

vi

vyi

vxi

y

θ

θiθ

θiθ!

"

#$

%

Active Figure 4.7 The parabolic path of a projectile that leaves the origin witha velocity vi. The velocity vector v changes with time in both magnitude anddirection. This change is the result of acceleration in the negative y direction.The x component of velocity remains constant in time because there is no accel-eration along the horizontal direction. The y component of velocity is zero atthe peak of the path.

A welder cuts holes through a heavymetal construction beam with a hottorch. The sparks generated in theprocess follow parabolic paths.

The

Tele

grap

h Co

lour

Lib

rary

/Get

ty Im

ages

Repeating with the y component and using yi ! 0 and ay ! " g, we obtain

(4.12)

Next, from Equation 4.11 we find t ! xf/(vi cos #i) and substitute this expression for tinto Equation 4.12; this gives

This equation is valid for launch angles in the range 0 $ #i $ %/2. We have left thesubscripts off the x and y because the equation is valid for any point (x, y) along thepath of the projectile. The equation is of the form y ! ax " bx 2, which is the equationof a parabola that passes through the origin. Thus, we have shown that the trajectory ofa projectile is a parabola. Note that the trajectory is completely specified if both the ini-tial speed vi and the launch angle #i are known.

The vector expression for the position vector of the projectile as a function of timefollows directly from Equation 4.9, with a ! g:

This expression is plotted in Figure 4.8, for a projectile launched from the origin, sothat ri ! 0.

The final position of a particle can be considered to be the superposition of the ini-tial position ri, the term vit, which is the displacement if no acceleration were present,and the term gt 2 that arises from the acceleration due to gravity. In other words, ifthere were no gravitational acceleration, the particle would continue to move along astraight path in the direction of vi. Therefore, the vertical distance gt 2 through whichthe particle “falls” off the straight-line path is the same distance that a freely falling ob-ject would fall during the same time interval.

In Section 4.2, we stated that two-dimensional motion with constant accelerationcan be analyzed as a combination of two independent motions in the x and y direc-tions, with accelerations ax and ay . Projectile motion is a special case of two-dimensional motion with constant acceleration and can be handled in this way, withzero acceleration in the x direction and ay ! " g in the y direction. Thus, when ana-lyzing projectile motion, consider it to be the superposition of two motions:

12

12

rf ! ri & vit & 12 gt 2

y ! (tan #i)x " ! g2v 2

i cos2 #i"x 2

yf ! vyit & 12ayt 2 ! (vi sin #i)t " 1

2gt 2

! PITFALL PREVENTION

4.2 Acceleration at theHighest Point

As discussed in Pitfall Prevention2.8, many people claim that theacceleration of a projectile atthe topmost point of its trajec-tory is zero. This mistake arisesfrom confusion between zerovertical velocity and zero acceler-ation. If the projectile were toexperience zero acceleration atthe highest point, then its veloc-ity at that point would notchange—the projectile wouldmove horizontally at constantspeed from then on! This doesnot happen, because the acceler-ation is NOT zero anywherealong the trajectory.

At the Active Figures linkat http://www.pse6.com, youcan change launch angle andinitial speed. You can also ob-serve the changing compo-nents of velocity along the tra-jectory of the projectile.

como se caracteriza a velocidade inicial, v0:

na horizontal? e na vertical ?

como se caracteriza o movimento: na vertical?

e na horizontal?

Movimento de Projéteis

Direção vertical - a intensidade e sentido do vector velocidade variam no tempo - esta variação resulta da aceleração da gravidade

Direção horizontal - o vector velocidade mantém-se constante - não há aceleração nesta direção

O lançamento de um projétil é uma

composição de dois movimentos

movimento uniforme, na

direção horizontal

movimento uniformemente

variado, na direção vertical

Movimento de Projéteis

Movimento de Projéteis Direção

horizontal (eixo xx’)

Direção vertical (eixo yy’)

ax = 0 ay = g

v0x = v0 cosα v0y = v0 senα

x = x0 + v0x t y= y0 + v0yt + 1/2 ayt2

Movimento de Projéteis

Trajetória dependente de

intensidade

Direção

velocidade inicial a=g

ângulo com a

horizontal

Movimento de Projéteis Lançamento horizontal

Existe velocidade inicial nas duas dimensões x e y?

v0x = v0 v0y = 0

Qual a velocidade em função do tempo ao longo da trajetória para as duas

dimensões x e y?

vx(t) = v0 vy (t) = -gt

Estratégias para resolver problemas de movimento

de projéteis:Definir o eixo de coordenadas e representar velocidade inicial em função de x e de y.

Estudar o movimento na componente horizontal - aplicando as leis do movimento uniforme

Estudar o movimento na componente vertical - aplicando as leis do movimento uniformemente acelerado

Problema:Um avião que se desloca na direção horizontal, 100 m acima do solo e com uma velocidade de 40 m/s, larga uma saco de mantimentos para dois exploradores que se encontram a trabalhar no local.

Qual será o alcance do saco relativamente ao ponto em que foi lançado?

90 C H A P T E R 4 • Motion in Two Dimensions

100 m

x

40.0 m/s

y

Figure 4.15 (Example 4.6) A package of emergency supplies isdropped from a plane to stranded explorers.

Example 4.7 The End of the Ski Jump

A ski-jumper leaves the ski track moving in the horizontal di-rection with a speed of 25.0 m/s, as shown in Figure 4.16.The landing incline below him falls off with a slope of 35.0°.Where does he land on the incline?

Solution We can conceptualize this problem based on obser-vations of winter Olympic ski competitions. We observe theskier to be airborne for perhaps 4 s and go a distance ofabout 100 m horizontally. We should expect the value of d,the distance traveled along the incline, to be of the same or-der of magnitude. We categorize the problem as that of a par-ticle in projectile motion.

To analyze the problem, it is convenient to select the be-ginning of the jump as the origin. Because vxi ! 25.0 m/sand vyi ! 0, the x and y component forms of Equation 4.9aare

(1)

(2) yf ! vyit " 12ayt2 ! #1

2(9.80 m/s2)t2

xf ! vxit ! (25.0 m/s)t

From the right triangle in Figure 4.16, we see that thejumper’s x and y coordinates at the landing point arexf ! d cos 35.0° and yf ! # d sin 35.0°. Substituting theserelationships into (1) and (2), we obtain

(3)

Solving (3) for t and substituting the result into (4), we findthat d ! 109 m. Hence, the x and y coordinates of the pointat which the skier lands are

To finalize the problem, let us compare these results to ourexpectations. We expected the horizontal distance to be onthe order of 100 m, and our result of 89.3 m is indeed on

# 62.5 myf ! #d sin 35.0$ ! #(109 m)sin 35.0$ !

89.3 mxf ! d cos 35.0$ ! (109 m)cos 35.0$ !

(4) #d sin 35.0$ ! #12(9.80 m/s2)t2

d cos 35.0$ ! (25.0 m/s)t

Example 4.6 The Stranded Explorers

A plane drops a package of supplies to a party of explorers,as shown in Figure 4.15. If the plane is traveling horizontallyat 40.0 m/s and is 100 m above the ground, where does thepackage strike the ground relative to the point at which it isreleased?

Solution Conceptualize what is happening with the assis-tance of Figure 4.15. The plane is traveling horizontallywhen it drops the package. Because the package is in free-fall while moving in the horizontal direction, we categorize

this as a projectile motion problem. To analyze the problem,we choose the coordinate system shown in Figure 4.15, inwhich the origin is at the point of release of the package.Consider first its horizontal motion. The only equation avail-able for finding the position along the horizontal directionis xf ! xi " vxit (Eq. 4.9a). The initial x component of thepackage velocity is the same as that of the plane when thepackage is released: 40.0 m/s. Thus, we have

If we know t, the time at which the package strikes theground, then we can determine xf , the distance the packagetravels in the horizontal direction. To find t, we use theequations that describe the vertical motion of the package.We know that, at the instant the package hits the ground, itsy coordinate is yf ! # 100 m. We also know that the initialvertical component of the package velocity vyi is zero be-cause at the moment of release, the package has only a hori-zontal component of velocity.

From Equation 4.9a, we have

Substitution of this value for the time into the equationfor the x coordinate gives

The package hits the ground 181 m to the right of the droppoint. To finalize this problem, we learn that an objectdropped from a moving airplane does not fall straight down.It hits the ground at a point different from the one right below the plane where it was released. This was an impor-tant consideration for free-fall bombs such as those used inWorld War II.

181 mxf ! (40.0 m/s)(4.52 s) !

t ! 4.52 s

#100 m ! #12(9.80 m/s2)t2

yf ! #12gt 2

xf ! (40.0 m/s)t