Lecture Notes 12.6

MOTION OF A PROJECTILE

Todays Objectives: Students will be able to: 1. Analyze the free-flight

motion of a projectile.

In-Class Activities: Reading Quiz

Applications

Kinematic Equations for

Projectile Motion

Concept Quiz

Group Problem Solving

Attention Quiz

READING QUIZ

1. The downward acceleration of an object in free-flight motion is

A) zero. B) increasing with time.

C) 9.81 m/s2. D) 9.81 ft/s2.

2. The horizontal component of velocity remains ________ during a free-flight motion.

A) zero B) constant

C) at 9.81 m/s2 D) at 32.2 ft/s2

APPLICATIONS

A good kicker instinctively knows at what angle, , and initial velocity, vA, he must kick the ball to make a field goal.

For a given kick strength, at what angle should the ball be kicked to get the maximum distance? Assuming level ground and eliminating t in the equations for x and y (setting y=0) we obtain x=R=(vA2/g)(2)(sin()cos()=(vA2/g)sin(2), dR/d=0=(2vA2/g )cos(). Therefore, R is maximum at =45.

APPLICATIONS (continued)

A basketball is shot at a certain angle. What parameters should the shooter consider in order for the basketball to pass through the basket?

Distance, speed, the basket location, anything else ? Shooting height and angle.

APPLICATIONS (continued)

A firefighter needs to know the maximum height, H, on the wall he could project water from the hose in the illustration above. What parameters would you program into a wrist computer to find the angle, , that he should use to hold the hose? See Example Problem III.

H = ?

MOTION OF A PROJECTILE (Section 12.6)

Projectile motion can be treated as two rectilinear motions, one in the horizontal direction experiencing zero acceleration and the other in the vertical direction experiencing constant acceleration (i.e., from gravity).

For illustration, consider the two balls on the left. The red ball falls from rest, whereas the yellow ball is given a horizontal velocity. Each picture in this sequence is taken after the same time interval. Notice both balls are subjected to the same downward acceleration since they remain at the same elevation at any instant. Also, note that the horizontal distance between successive photos of the yellow ball is constant since the velocity in the horizontal direction is constant.

KINEMATIC EQUATIONS: HORIZONTAL MOTION

Since ax = 0, the velocity in the horizontal direction remains constant (vx = vox) and the position in the x direction can be determined by:

x = xo + (vox) t Eqn. 1 Why is ax equal to zero (assuming movement through the air)? The assumption, though correct only for a dimensionless particle, is that we neglect air resistance or drag force and, hence, no acceleration (deceleration) in the direction of motion.

KINEMATIC EQUATIONS: VERTICAL MOTION

Since the positive y-axis is directed upward, ay = g. Application of the constant acceleration equations yields:

vy = voy g t Eqn. 2

y = yo + (voy) t g t2 Eqn. 3 vy2 = voy2 2 g (y yo) Eqn. 4

For any given problem, only two of these three equations can be used. Why? Only two independent equations

EXAMPLE I Given: vo and Find: The equation that defines

y as a function of x. Plan: Eliminate time from the

kinematic equations.

Solution: Using vx = vo cos and vy = vo sin We can write: x = (vo cos )t or y = (vo sin ) t g (t)2

t = x vo cos

y = (vo sin ) { } { }2 x g x vo cos 2 vo cos

By substituting for t:

EXAMPLE I (continued)

The above equation is called the path equation which describes the path of a particle in projectile motion. The equation shows that the path is parabolic.

Simplifying the last equation, we get:

y = (x tan) g x2

2vo2 (1 + tan2)

EXAMPLE II

Given: Projectile is fired with vA=150 m/s at point A.

Find: The horizontal distance it travels (R) and the time in the air.

Plan:

Establish a fixed x, y coordinate system (in this solution, the origin of the coordinate system is placed at A). Apply the kinematic relations in x- and y-directions.

EXAMPLE II (continued)

Solving for tAB first, tAB = 19.89 s. (Use quadratic equation) Then, R = 120 tAB = 120 (19.89) = 2387 m

Solution: 1) Place the coordinate system at point A. Then, write the equation for horizontal motion. + xB = xA + vAx tAB

where xB = R, xA = 0, vAx = 150 (4/5) m/s Range, R will be R = 120 tAB

2) Now write a vertical motion equation. Use the distance equation. + yB = yA + vAy tAB 0.5 g tAB2 where yB = 150, yA = 0, and vAy = 150(3/5) m/s We get the following equation: 150 = 90 tAB + 0.5 ( 9.81) tAB2

CONCEPT QUIZ

1. In a projectile motion problem, what is the maximum number of unknowns that can be solved?

A) 1 B) 2 C) 3 D) 4

2. The time of flight of a projectile, fired over level ground, with initial velocity Vo at angle , is equal to?

A) (vo sin )/g B) (2vo sin )/g C) (vo cos )/g D) (2vo cos )/g Set y = y0 in y equation and solve for t or recognize that the final vy is -voy and solve for t using the vy equation.

Three independent equations & three unknowns: x = xo + (vox) t , vy = voy g t, and y = yo + (voy) t g t2

EXAMPLE III Given: A firefighter equipped with a

wrist calculator must stand 30 feet from a burning building as shown to the right. He knows that the velocity of the water leaving the nozzle is 48 ft/s.

Find: The maximum height h on the wall to which the firefighter can project water from the hose and the angle determined by wrist calculator at which the nozzle must be held to achieve this maximum height.

Plan: Assuming we are not concerned about the angle at which the water strikes the building, the solution is to maximize the height of the water at a distance of 30 ft from the nozzle. Note that time is not important. Therefore, maximizing y with respect to is the plan.

EXAMPLE III (continued) Solution:

From Example I we found the following path equation:

y = (x tan) g x2

2vo2 (1 + tan2)

We know from calculus that the maximum (or minimum) of a function relative to its variable argument can be obtained by differentiating the function with respect to its variable argument and setting the result equal to zero. This is true, of course, since a maxima or minima is achieved when the slope of the function is zero. To this end, we differentiate Eq. (1) with respect to and set the result equal to zero. Note, for this problem x, vo, and g are constants.

Eq. (1)

EXAMPLE III (continued)

EXAMPLE III (continued)

EXAMPLE III (continued)

GROUP PROBLEM SOLVING

Given: A skier leaves the ski jump ramp at A = 25o and hits the slope at B.

Find: The skiers initial speed vA. Plan:

Establish a fixed x,y coordinate system (in this solution, the origin of the coordinate system is placed at A). Apply the kinematic relations in x and y-directions.

x y

GROUP PROBLEM SOLVING (continued)

Motion in x-direction: Using xB = xA + vox(tAB) => (4/5)100 = 0 + vA (cos 25) tAB

= tAB= 80

vA (cos 25) 88.27 vA

vA = 19.42 m/s

64 = 0 + vA(sin 25) { } 88.27

vA (9.81) { }2 88.27 vA

Motion in y-direction:

Using yB = yA + voy(tAB) g(tAB)2

Solution:

yB = -[(3/5)100 + 4] = 64

ATTENTION QUIZ

1. A projectile is given an initial velocity vo at an angle above the horizontal. The velocity of the projectile when it hits the slope is _____A______ the initial velocity vo. A) less than B) equal to C) greater than D) None of the above.

2. A particle has an initial velocity vo at angle with respect to the horizontal. The maximum height it can reach is when

A) = 30 B) = 45

C) = 60 D) = 90 v2y = v2oy 2g(y-yo) @ max y vy = 0 and 2g(y-yo) = (vosin )2 which is max. @ = 90

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