Copyright Sautter 2011

Please let me promote some of my other work on the next slide and

then the selected presentation will start. Thank you.

Walt S.

I have written six books:

"Sticks - A Golfer's Tale" - A hacker's dream comes true.

"Fish Farm" - Revenge of the old people.

"Coach" - A mystery set a rural, bigoted, nineteen fifties football town.

"The Three Dollar Phoenix" - A mystery set in Newark, New Jersey in the 1970s.

"The Divine Comedy MMIX" - A humorous play about Jesus returning.

"The Blood of Judas" - A horror story of revenge set in Nazi Germany.

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Walt Sautter - wsautter@optonline.net

PS - Lots more stuff besides booksat: wsautter.com

Motion in Two Dimension - Projectiles• Projectile motion involves object that move up or down and

right or left simultaneously. A ball thrown into the air at an angle is a common example as is water sprayed from a garden hose or a bullet fired at a target.

• Any object once released from its accelerating force is accelerated downward by gravity at all times during its flight. This causes a continual variation in its vertical velocity while its horizontal velocity remains unaffected.

• It discussing projectile motion, the horizontal and vertical components are treated independently. The combined effect of these two motions on the object give its parabolic flight path. When objects are projected from a specific horizontal elevation and land at the same horizontal elevation, the parabolic path of the object is symmetrical.

• The next slide shows a projectile path and key points during the flight in terms of displacements(Sy and Sx), velocities (Vyand Vx) and accelerations (Ay and Ax) ( g, of course, refers to gravity)

JUST AFTERFIRING

Sy =0, Sx = 0Vy = +max

Vx = constantAy = g, Ax =0

AT HIGH POINTSy = max, Sx = ½ maxVy = 0, Vx = constant

Ay = g, Ax =0

JUST BEFORE LANDING

Sy = 0, Sx = maxVy = -max

Vx = constantAy = g, Ax =0

CLICKHERE

A

X COMPONENT

Y COMPONENT

X COMPONENT

Y COMPONENTB

Y COMPONENT

X COMPONENT

C

Sy

Time

Vy inst = slope of tangents

VyTime

Vy = 0

Ay inst = slope of tangents

Ay TimeAy inst = - 9.8 meters / sec2

Vo

Sx

Path of objectWithout gravity

Sy = actual height of object

Distance fallen dueto gravity (1/2 g t2)

Vertical height if gravity did not act on the object (Vo sin t)

Sy = height without effects of gravity – distance fallen due to gravitySy = Vo sin t + ½ g t2 (the value of gravity is negative)

Horizontal distance traveled by projectile Sx = Vo cos t

Actual projectilepath

ObjectProjected at

Angle

Water spraying froma hose is a common

example of projectilemotion

The vertical motion ofthe water is accelerated

by gravity. The horizontalmotion is constant

velocity and is unaffectedby gravity!

Vy = Vo sin + gtVx = Vo cos

Vertical displacements are equal for projectiles and dropped objects however horizontal displacements are greater for projectiles.

Vertical acceleration for both dropped objects and projectiles is that of gravity (-32 ft/s2, -9.8 m/s2). They both hit the ground at the exact

same time but, of course, the projectile is further away !

Accelerated bygravity

Projectile Vx > 0Constant velocity

DroppedObjectVx = 0

Horizontal component of projectile motion(Constant velocity – no acceleration)

Vertical component of projectile motion(Accelerated by gravity)

Vy = + max

Vy = - max

Vy = 0

VerticalDisplacement

HorizontalDisplacement

VerticalVelocity

HorizontalVelocity

VerticalAcceleration

HorizontalAcceleration

Maximum Height

Maximum HorizontalDistance (Range)

hmax

range

When an object is dropped at the exact same time a projectile is fired at the

falling object, aimingdirectly at the object

always insures a direct hit.Why ?

Because the object and the projectile once fired,

are in both in free fall !

Solving Projectile ProblemsA ball is thrown horizontally at 12 m/s from a building 30 meters high. (a) How long will is be in the air? (b) How far from the base

of the building will it land?

“horizontally” means the angle of projection is 0 degrees

In part (a) we are asked to find time when the vertical distance is – 30 meters (negative means below the point of release).

In part (b) we are asked to find horizontal distance

• (a) Using the equation shown and inserting –30 for vertical distance, 12 for the original velocity, 0 degrees for the angle and – 9.8 for gravity (MKS) we get:

-30 = 12 (sin 0)t + ½ (-9.8)t2, solving for t gives 2.47sec

(b) Using the equation shown and inserting 12 for

the original velocity, 0 degrees for the angle and

2.4 seconds for time found in part (a) we get

Vx = 12 (cos 0) 2.47 = 29.6 meters

Vo = 12 m/s

30 m

• The object is projected at + 300. Vo = 12 ft/s and we are asked in part (a) to find the time when Sy = - 300 ft.

• In part (b) we are asked to find the horizontal distance (Sx).

Solving Projectile ProblemsA ball is thrown upward at a 300 angle, at 12 ft/s from a building 300 feet high. (a)How long will is be in the air? (b) How far from the base of the building will it land?

Vo = 12 m/s

300 m

300

(a) using the equation shown we insert –300 for Sy, 12 for Vo 300 for the angle and – 32 ft/s2 for gravity (ENG) and get –300 = 12(sin 300 ) t + ½ (-32) t2 which gives –300 = 6 t – 16 t2 or 16 t2 –6 t - 300 = 0 (a quadratic). Using a = +16, b = - 6 and c = - 300 we insert these values in the quadratic equation find t = -3.21 or + 5.46The negative t means the ball hits the ground before it is thrown in is therefore obviously wrong. t = +5.46 sec

(b) Using the equation shown and the time value from part (a) we get Vx = 12 (cos 300) 5.46 = 56.7 feet

Solving Projectile ProblemsA projectile is shot on level ground at a 45 degree angle with a velocity of 20

ft/s. (a) How high will it travel? (b) How far will it go?

• Vo = 20, the angle = 450 and in part (a) we are asked to find hmax

• In part (b) we are asked to find the range ( R)

hmax

R

(a) using the equation shown we insert 20 for Vo 450 for the angle and – 32 ft/s2 for gravity (ENG) and get hmax = - ((20)2 (sin 450)2 )/ (2 (-32)) which gives (400 x (0.707)2) / -64 . Maximum height = 3.13 feet.

(b) Using the equation shown we get R = - ((20)2 x (sin 2 x 450)) / - 32 = 12.5 feet

Solving Projectile ProblemsA ball is throw downward at an angle of 300 with a velocity of 10 m/s from a

building 40 meters high. What is the velocity of the ball when it hits the ground?

Vo = -10 m/s

40 m

-300

Downward = - 300

We are asked to find the velocity (a vector quantity - direction counts) when the object hits the ground. At this point it is moving both horizontally and vertically.

We will first find the Vx value and the Vy value and add them using vectors.

Using the given equations, Vo = 10 m/s, g = - -9.8 m/s2 (MKS) and Sy = - 40 m we must first find time (t). -40 = 10 sin (-300) t + (-9.8) t2 , 9.8 t2 + 5 t – 40 = 0. Solving the quadratic formula with a = 9.8, b = 5 and c = -40 we get t = - 2.29 or + 1.78 seconds. Of course, the + 1.78 is the correct answer.

To find the vertical velocity, Vy = 10 (sin – 300) + (-9.8) 1.78, Vy = -22.5 m/s

To find the horizontal velocity, Vx = 10 (cos –300) 1.78 = +15.4 m/s

Previous Problem (continued)

• When the object hits the ground it is moving both vertically under the influence of gravity and horizontally at constant velocity (unaccelerated).

Vy = -22.5 m/s

Vx = +15.4 m/s

Vresulting

• To find V resulting we use vector addition (the Pythagorean Theorem), Vr = ((-22.5)2 + (15.4)2)1/2 , Vr = 27.3 m/s

• Vector values such as velocity requires a direction also. Using the inverse tangent we get, tan –1 (15.4 / 22.5 ) = 34.4 0

• The velocity at which the object hits the ground is 27.3 m/s and an angle of 34.4 degrees below the horizontal.

• A projectile is shot on level ground at a 45 degree angle with a velocity of 20 ft/s. When will its velocity be –5 ft/s?

Solving Projectile Problems

Vo = 20

450

V = - 5, t = ?

• Vy = -5 ft/s, Vo = 20 ft/s and the angle of projection = 450. Using the given equation, -5 = 20 (sin 450) + (- 32 ) t

• t = 0.598 seconds

Check - At the highest point, Vy =0 and 0 = 20 (sin 450) + (-32) t = 0.442 seconds at the highest point and when descending t must be larger than 0.442 and smaller than 0.884 (2 x 0.442), the time when the object lands. The answer is between these two values!

A ball is thrown horizontally at 12 m/s.(a) What is its velocity after 2.0 seconds?

(A) 12 m/s (B) 19.6 m/s (C) 23 m/s (D) 31.6 m/s

ClickHereFor

answers

A ball is thrown at 400 above the horizontal at 4.0 m/s.What is its horizontal velocity after 0.50 seconds?

(A) 2.6 m/s (B) 3.1 m/s (C) 3.4 m/s (D) 5.5 m/s

What is the vertical velocity in the previous problemAfter 0.50 seconds ?

(A) 2.6 m/s (B) 3.1 m/s (C) 4.9 m/s (D) 7.5 m/s

What must be the original velocity of a projectile in order to reach a target 90 km away ?

(A) 320 m/s (B) 939 m/s (C) 882 m/s (D) none

Gravity is its acceleration

It is -9.8 m /s2

(MKS units)

A ball is thrown at a 300 angle at 10 m/s. How faron the horizontal will it land ?

(A) 9.8 m (B) 0.88 ft (C) 8.8 m (D) 10.2 m

Hint: useThe vector

Sum of Vx + Vy

Hint: maximumhorizontal

distance occursat an angle of 450

Real Projectile Motion

• In actual free fall situations objects reach a terminal velocity and do not accelerate at a constant rate throughout the entire fall.

• The buoyant force to the air acts to reduce the acceleration caused by gravity. The degree to which this retarding force acts, depends on several variables.

• As the density of the air increases, buoyancy increases. Also, increased surface area increases the effects of buoyancy and causes a decrease in terminal velocity.

• The shape and aerodynamic properties of the object also effect terminal velocity.

• Most importantly, as the velocity of the falling body increases, the force opposing the fall increases, thus at a specific velocity the force of gravity and opposing force become equal and terminal velocity is reached.