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Physics Coach Taylor Broad Run High School 2010-2011.

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Presentation on theme: "Physics Coach Taylor Broad Run High School 2010-2011."— Presentation transcript:


2 Physics Coach Taylor Broad Run High School 2010-2011

3  Recognize that the vertical and horizontal motions of a projectile are independent.  Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, then determine the range.  Explain how the shape of trajectory of a moving object depends upon the frame of reference from which it is observed.

4  Pro – jec - tile : a body projected or impelled forward, as through the air.

5  Dropped straight down.  Thrown straight up.  Thrown with a horizontal and vertical velocity.


7  Horizontal motion does NOT affect the vertical motion. /gravitational_acceleration/horizontal_and_vertical_ball_drop.gif

8  Constant Horizontal Motion Constant Horizontal Motion  Horizontal motion is constant because it is initiated by a contact force.  Vertical motion is not constant because it is caused by a long-range force.

9  Horizontal and vertical arrows indicate respective velocities: v x & v y.  The velocity vector is the combined vector of the x and y components.

10  Projectiles launched horizontally have no initial vertical velocity.  Their vertical motion is identical to that of a dropped object.  Downward velocity increases regularly because of acceleration due to gravity.


12  When a projectile is launched at an angle, the initial velocity has vertical and horizontal components.  Starts with highest vertical velocity, reaches its peak where velocity is zero, and descends with increasing speed.

13  Maximum height: height of the projectile when the vertical velocity is zero and the projectile has only its horizontal velocity component.  Range, R: horizontal distance the projectile travels.  Flight-time: time the projectile is in the air.


15  Air resistance affects projectiles.  Golf ball dimples.  Spin of a Baseball due to laces.  Frisbee.  Boomerang.

16 A red marble is dropped off a cliff at the same time a black one is shot horizontally. At any point in time the marbles are at the same height, i.e., they’re falling down at the same rate, and they hit the ground at the same time. Gravity doesn’t care that the black ball is moving sideways; it pulls it downward just the same. Since gravity can’t affect horiz. motion, the black particle continues at a constant rate. With every unit of time, the marbles’ vertical speed increases, but their horiz. speed remains the same (ignoring air resistance). continued on next slide

17 Gravity’s downward pull is independent of horiz. motion. So, the vertical acceleration of each marble is - g (for the whole trip), and the sideways acceleration of each is zero. (Gravity can’t pull sideways). Whatever horiz. velocity the black one had when shot is a constant throughout its trip. Only its vertical velocity changes. (A vertical force like gravity can only produce vertical acceleration.) 9.8 m/s 2

18 continued on next slide t = 0 t = 1 t = 2 t = 3 t = 4 v y = 1 v y = 2 v y = 3 v y = 4 v y = 0 If after one unit of time the marbles have one unit of speed downward, then after two units of time they have two units of speed downward, etc. This follows directly from v f = v 0 + a t. Since v 0 = 0, downward speed is proportional to time. Note: The vectors shown are vertical components of velocity. The shot marble has a horizontal component too (not shown); the dropped one doesn’t.

19 Since the shot black marble experiences no horiz. forces (ignoring air), it undergoes no horiz. acceleration. Therefore, its horiz. velocity, doesn’t change. So, the horiz. vector has a constant magnitude, but the vertical vector gets longer. The resultant (the net velocity vector in blue) gets longer and points more downward with time. When t = 0, v = v x for the shot marble. v = v y for the dropped marble for the whole trip. continued on next slide v x = v vxvx vxvx vxvx vxvx v v v v y t = 0 t = 1 t = 2 t = 3 t = 4

20 The trajectory of any projectile is parabolic. (We’ll prove this later.) If its initial velocity vector is horizontal, as with the black marble, the launch site is at the vertex of the parabola. The velocity vector at any point in time is tangent to the parabolic trajectory. Moreover, velocity vectors are always tangent to the trajectory of any moving object, regardless of its shape. v v v continued on next slide

21 x = 1  y = 1  y = 3  y = 5  y = 7 x = 2 x = 3 x = 4 continued on next slide The vertical displacements over consecutive units of time show the familiar ratio of odd numbers that we’ve seen before with uniform acceleration. Measured from the starting point, the vertical displacements would be 1, 4, 9, 16, etc., (perfect squares), but the horiz. displacements form a linear sequence since there is no acceleration in that direction. t = 0 t = 1 t = 2 t = 3 t = 4

22 A rifle is held perfectly horizontally 1.5 m over level ground. At the instant the trigger is pulled, a second bullet is dropped from the tip of the barrel. The muzzle velocity of the gun is 80 m/s. 1. Which bullet hits the ground first? answer: 2. How fast is each bullet moving after 0.3 s ? answer: dropped bullet after 0.3 s fired bullet after 0.3 s 80 m/s They hit at same time. vyvy vyvy Use v f = v 0 + a t and use vertical info only: v 0 = 0, a = -9.8 m/s 2, and t = 0.3 s. We get v y in the pic for each bullet is -2.94 m/s. Using the Pythagorean theorem for the fired bullet we get 80.054 m/s in a direction tangent to its path. continued on next slide

23 80 m/s 1.5 m 3. How far away does the fired bullet land (its range)? answer: The first step is to find the its hang time. This is the same hang time as the dropped bullet. Use  y = v 0 t + 0.5 a t 2 with only vertical data: -1.5 = (0) t + (0.5) (-9.8) t 2. So, t = 0.5533 s. The whole time the bullet is falling it’s also moving to the left at a constant 80 m/s. Since horizontally v is constant, we use d = v t with only horiz. info: d = (80 m/s) (0.5533 s) = 44.26 m. Note: When a = 0,  x = v 0 t + 0.5 a t 2 breaks down to d = v t.

24 Now let’s find range of a projectile fired with speed v 0 at an angle . Step 1: Split the initial velocity vector into components. v 0 cos  v 0 sin  v0v0  continued on next slide

25   The projectile’s speed is the same at points directly across the parabola (at the same vertical position). The angle is the same too, but with opposite orientation. Horizontal speeds are the same throughout the trajectory. Vertical speeds are the same only at points of equal height.  The vert. comp. shrinks then grows in opposite direction at a const. rate (- g ). The resultant velocity vector’s orientation and magni- tude changes, but is always tangent. The horiz. comp. doesn’t change. At the peak, the horiz. comp. equals the resultant velocity vector.

26 t = 0 t = 10 t = 20 t = 15 t = 5 t = 3 t = 17 Over level ground, the time at the peak is half the hang time. Notice the symmetry of times at equal heights relative to the 10 unit mark. The projectile has covered half its range when it has peaked, but only over level ground. Note: near the peak the object moves more slowly than when lower to the ground. It rises 3/4 of its max height in only 1/2 of its rising time. (See if you can prove this for an arbitrary launch velocity.)

27 Each initial velocity vector below has the a different magnitude (speed) but each object will spend the same time in the air and reach the same max height. This is because each vector has the same vertical component. The projectiles will have different ranges, however. The greater the horizontal component of initial velocity, the greater the range.

28 Over level ground at a constant launch speed, what angle maximizes the range, R ? First consider some extremes: When  = 0, R = 0, since the object is on the ground from the moment it’s launched. When  = 90 , the object goes straight up and lands right on the launch site, so R = 0 again. The best angle is 45 , smack dab between the extremes. 45  Here all launch speeds are the same; only the angle varies. 38  76  proof on next slide

29 When fired from a cliff, or from below ground, a projectile doesn’t attain its max range at 45 . 45  is only the best angle when a projectile is fired over level ground. When fired from a cliff, a projectile attains max range with a launch angle less than 45  (see next slide). When fired from below ground, a projectile attains max range with a launch angle greater than 45 .

30 45  < 45  If ground were level, the 45  launch would win. Because the < 45  parabola is flatter it eventually overtakes 45  parabola. Launch speeds are the same.

31 I. Law of Inertia II. F=ma III. Action-Reaction

32 While most people know what Newton's laws say, many people do not know what they mean (or simply do not believe what they mean).

33  1 st Law  1 st Law – An object at rest will stay at rest, and an object in motion will stay in motion at constant velocity, unless acted upon by an unbalanced force.  2 nd Law  2 nd Law – Force equals mass times acceleration.  3 rd Law  3 rd Law – For every action there is an equal and opposite reaction.

34 An object at rest will stay at rest, and an object in motion will stay in motion at constant velocity, unless acted upon by an unbalanced force.

35  Inertia is the tendency of an object to resist changes in its velocity: whether in motion or motionless. These pumpkins will not move unless acted on by an unbalanced force.

36  Once airborne, unless acted on by an unbalanced force (gravity and air – fluid friction), it would never stop!

37  Unless acted upon by an unbalanced force, this golf ball would sit on the tee forever.

38 Why then, do we observe every day objects in motion slowing down and becoming motionless seemingly without an outside force? It’s a force we sometimes cannot see – friction.

39 Objects on earth, unlike the frictionless space the moon travels through, are under the influence of friction.

40  There are four main types of friction:  Sliding friction: ice skating  Rolling friction: bowling  Fluid friction (air or liquid): air or water resistance  Static friction: initial friction when moving an object What is this unbalanced force that acts on an object in motion?

41 Slide a book across a table and watch it slide to a rest position. The book comes to a rest because of the presence of a force - that force being the force of friction - which brings the book to a rest position.

42  In the absence of a force of friction, the book would continue in motion with the same speed and direction - forever! (Or at least to the end of the table top.)

43 Don’t let this be you. Wear seat belts. Because of inertia, objects (including you) resist changes in their motion. When the car going 80 km/hour is stopped by the brick wall, your body keeps moving at 80 m/hour.


45 The net force of an object is equal to the product of its mass and acceleration, or F=ma.

46  When mass is in kilograms and acceleration is in m/s/s, the unit of force is in newtons (N).  One newton is equal to the force required to accelerate one kilogram of mass at one meter/second/second.

47  How much force is needed to accelerate a 1400 kilogram car 2 meters per second/per second?  Write the formula  F = m x a  Fill in given numbers and units  F = 1400 kg x 2 meters per second/second  Solve for the unknown  2800 kg-meters/second/second or 2800 N

48 If mass remains constant, doubling the acceleration, doubles the force. If force remains constant, doubling the mass, halves the acceleration.

49  We know that objects with different masses accelerate to the ground at the same rate.  However, because of the 2 nd Law we know that they don’t hit the ground with the same force. F = ma 98 N = 10 kg x 9.8 m/s/s F = ma 9.8 N = 1 kg x 9.8 m/s/s

50  1. What acceleration will result when a 12 N net force applied to a 3 kg object? A 6 kg object?  2. A net force of 16 N causes a mass to accelerate at a rate of 5 m/s 2. Determine the mass.  3. How much force is needed to accelerate a 66 kg skier 1 m/sec/sec?  4. What is the force on a 1000 kg elevator that is falling freely at 9.8 m/sec/sec?

51  1. What acceleration will result when a 12 N net force applied to a 3 kg object? 12 N = 3 kg x 4 m/s/s  2. A net force of 16 N causes a mass to accelerate at a rate of 5 m/s 2. Determine the mass. 16 N = 3.2 kg x 5 m/s/s  3. How much force is needed to accelerate a 66 kg skier 1 m/sec/sec? 66 kg-m/sec/sec or 66 N  4. What is the force on a 1000 kg elevator that is falling freely at 9.8 m/sec/sec?  9800 kg-m/sec/sec or 9800 N


53  For every action, there is an equal and opposite reaction.

54 According to Newton, whenever objects A and B interact with each other, they exert forces upon each other. When you sit in your chair, your body exerts a downward force on the chair and the chair exerts an upward force on your body.

55 There are two forces resulting from this interaction - a force on the chair and a force on your body. These two forces are called action and reaction forces.

56  Consider the propulsion of a fish through the water. A fish uses its fins to push water backwards. In turn, the water reacts by pushing the fish forwards, propelling the fish through the water.  The size of the force on the water equals the size of the force on the fish; the direction of the force on the water (backwards) is opposite the direction of the force on the fish (forwards).

57 Flying gracefully through the air, birds depend on Newton’s third law of motion. As the birds push down on the air with their wings, the air pushes their wings up and gives them lift.

58  Consider the flying motion of birds. A bird flies by use of its wings. The wings of a bird push air downwards. In turn, the air reacts by pushing the bird upwards.  The size of the force on the air equals the size of the force on the bird; the direction of the force on the air (downwards) is opposite the direction of the force on the bird (upwards).  Action-reaction force pairs make it possible for birds to fly.

59  The baseball forces the bat to the left (an action); the bat forces the ball to the right (the reaction).

60  Consider the motion of a car on the way to school. A car is equipped with wheels which spin backwards. As the wheels spin backwards, they grip the road and push the road backwards.

61 The reaction of a rocket is an application of the third law of motion. Various fuels are burned in the engine, producing hot gases. The hot gases push against the inside tube of the rocket and escape out the bottom of the tube. As the gases move downward, the rocket moves in the opposite direction.

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