Projectiles

1. Projectiles A projectile is any object projected by some means that continues in motion by its own inertia.

2. Outline Projectile motion Worked examples Practical and theoretical examples

3. Projectile components Projectile motion is treated as two separate problems (horizontal and vertical). For most problems a variable is solved for in one direction, and then the value is carried into the second direction to solve for the desired quantity

4. Motion of projectiles

5. Horizontally Launched Projectiles

6. Non-Horizontally Launched Projectiles

7. Both vertical and horizontal components of projectile motion

9. Vectors and projectiles

10. Vectors and projectiles

11. Description of Vertical Motion for an Object Thrown Upward: Upward Path Velocity is upward Acceleration due to gravity is downward Since acceleration and velocity are in opposite directions, the object slows down as it moves upward

12. At Maximum Height Velocity has slowed to zero Acceleration is still present! Acceleration is changing the direction of motion

13. Downward Path Velocity is downward Acceleration due to gravity is downward Since acceleration and velocity are in the same directions, the object speeds up as it falls downward

14. Points to remember  Our standard convention is that upward is positive. Any quantity (displacement, velocity, acceleration) pointing upward has a positive sign. Any quantity (displacement, velocity, acceleration) pointing downward has a negative sign. Acceleration (g) is due to the gravitational attraction of the earth Acceleration always acts downward. (Therefore it is negative) Acceleration is always equal to -9.81 m/s2 for a freely falling object At maximum height velocity is zero At maximum height acceleration is still -9.81m/s2, but is causing the direction to change while the velocity remains zero

15. Equations Equations for constant acceleration (or laws of uniformly accelerated motion). These laws are used for obtaining the vertical or horizontal components of projectile motion but they are not applicable for the resultant motion of a projectile. v = vo + at d = � (v + vo)t d = vot + � at2 v2 = vo2 + 2ad Where V is final velocity, Vo is initial velocity and d is distance, a is acceleration and t is time. It is important to note that horizontal acceleration is always zero and the vertical acceleration is -9.81 m/s2. At the apex of a projectile�s trajectory the vertical component of velocity is zero. Further, the time a projectile takes to reach the apex is one half of the total flight time if the projection and landing heights are equal. Therefore final velocity will be zero at the apex allowing: 0 = vo + at

16. Equations for the resultant motion of the projectile Sin ? and cos ? are trigonometric functions where horizontal velocity is the angle in radians times cosine, and the vertical velocity is sine times the angle in radians. vx0=v0 cos h ?, vy0=v0 sin h ? Horizontal distance or range = v2 sin 2? / gravity ? is the angle measured in radians. v is average horizontal velocity. Uneven horizontal ranges Uneven horizontal range = v2 sin ? cos ? + vcos ? v(vsin?)2 + 2gh / gravity Vertical time of flight Time of flight = 2 vsin? / g For differing start and finish heights Time for flight = vsin? + v(vsin?)2 + 2gh g is gravity and h is height of take off minus height of finish

17. Note: The time of the flight depends solely upon the initial vertical velocity. For differing start heights the change in vertical velocity and height of release will effect time of flight. For horizontal range for even heights the speed of release and angle of release will give greatest distance (45 degrees optimal angle of release). For uneven take off and finish heights the speed of release, angle of release and height of release will all affect horizontal range.

18. In the case of uneven heights for take off and landing the following are true. The optimum angle of release is always less than 45 degrees. For any given height of release, the greater the speed of release, the more closely the optimum angle approaches 45 degrees. For any given speed of release, the greater the height of release, the less is the optimum angle. Equal increases in either height of release or speed of release do not yield consistently equal changes in the optimum angle or the resulting distance.

19. Solving projectile motion problems Choose an origin for your x and y coordinates and make a sketch. Write down what you know and what you do not know. For convenience sake, make a table with horizontal information on one side and vertical information on the other side. Write down separate equations for the horizontal and vertical motion. Form a strategy for using your equations and your known information to find what you need. Check that your answer makes sense and the units are correct.

20. Example: vo = 40.0 m/s theta = 35 degrees

21. Components of original velocity The usual first step in this investigation is to find the x and y components for the original velocity. General: X component of original velocity: vox = vo cos (theta) Y component of original velocity: voy = vo sin (theta) Example: In the x direction: vox = vo cos (theta) vox = (40.0 m/s)(cos(35 degrees)) vox = (40.0)(0.8191) vox = 32.76 vox = 32.8 m/s In the y direction: voy = vo sin (theta) voy = (40.0 m/s)(sin(35 degrees)) voy = (40.0)(0.5735) voy = 22.94 voy = 22.9 m/s

22. How much time passes until the projectile is at the top of its trajectory? At the top of the trajectory the y, or upward, velocity of the projectile will be 0.0 m/s. The object is still moving at this moment, but its velocity is purely horizontal. At the top it is not moving up or down, only across.

23. Solution General: We can use the following kinematics equation: vf = vo + at Subscript it for y: vfy = voy + ayt Solve it for t: t = (vfy - voy) / ay Plug in 0.0 m/s for vfy: t = (0.0 m/s - voy) / ay If the original y velocity and the y acceleration, i. e., the acceleration due to gravity, are plugged into the above equation, it will solve for the amount of time that passes from the moment of release to the moment when the projectile is at the top of its flight.

24. Example: Start with: t = (vfy - voy) / ay Plug in 0.0 m/s for vfy: t = (0.0 m/s - voy) / ay Plug in values for voy and ay: t = (0.0 m/s - 22.9 m/s) / - 9.8 m/s2 t = -22.9 / -9.8 t = 2.33 t = 2.3 s In this example 2.3s of time passes while the projectile is rising to the top of the trajectory.

25. How high does the projectile rise? General: Here is the displacement formula: d = vot + 0.5at2 We must think of this displacement in the y direction, so we will subscript this formula for y: dy = voyt + 0.5ayt2 If now we plug in the half time of flight, which was found above, we will solve for the height of the trajectory, since the projectile is at its maximum height at this time.

26. Example: Starting with: dy = voyt + 0.5ayt2 Then plugging in known values: dy =�(22.9 m/s)(2.33 s) + (0.5)(-9.8 m/s2)(2.33 s)2 dy = 53.35 - 26.60 dy = 26.75 dy = 27 m

27. How much time passes until the projectile strikes the ground? General: With no air resistance, the projectile will spend an equal amount of time rising to the top of its projectile as it spends falling from the top to the ground. Since we have already found the half time of flight, we need only to double that value to get the total time of flight. Example: t = 2(2.33 s) t = 4.66 t = 4.7 s This is the total time of flight.

28. How far away does the projectile land from its starting point? General: Let us start with the general displacement formula: d = vot + 0.5at2 Since we are working in the x direction, we should subscript this equation for x: dx = voxt + 0.5axt2 Now, since the acceleration in the x direction is 0.0 m/s2, the second term in the above equation drops out, and we are left with: dx = voxt The velocity in the x direction does not change. The projectile maintains its original x velocity throughout its entire flight. So, the original x velocity is the only x velocity the projectile will have. We could, therefore, think of the last equation as: dx = vxt If we plug in the original x velocity for vx and the total time of flight for t, we will solve for the horizontal displacement, or range, of the trajectory.

29. Example: As shown in the general section above, start with: dx = vxt Plug in values. Remember that the x velocity is constant and always equal to its original value and that the time here is the total time of flight. dx = (32.8 m/s)(4.66 s) dx = 152.84 dx = 150 m

30. Example problem An archer shoots an arrow at a wall that is a horizontal distance of 50 m away. The initial velocity of the arrow is 100 m/s in a horizontal direction. The arrow begins at a height of 10 m above the ground. How far above the ground will the arrow be when it hits the wall?

31. Throw and push �like movements Throwing is the act of propelling an object through the air by means of the action of the body, climaxed by the arm and hand. The release is through extension or by whirling motion of the upper limb and parts of the body assisting in the action.

32. Throw and push-like movement patterns The aims of throwing / pushing in sport are: 1) To project the object the greatest vertical or horizontal distance 2) To project an object primarily for accuracy -with the speed of projection enhancing the projectiles effectiveness.

33. Terminology Constraint determines the variables that occur in time and space such as mass of the equipment, strength and ability of the performer. Open and closed kinetic chain movements determine whether the segment or joint is free to move in space or whether the end of the segment is in contact with another body.

34. Throw-like patterns: sequential segmental rotations Throw-like patterns are movements used to project an object that is allowed to lag behind the proximal segments that have finished their backswings and are now moving forward. Note: That the pectoralis major and latissimus dorsi are actually arm muscles, in terms of action, even though they are located in the thorax and back, they attach to the humerus in the bicipital groove. Many arm muscles are barticular; they move two joints in the same direction at the same time.

35. The open kinetic link model The kinetic link model is a segmental link system that aims to produce high distal end velocity. The velocity of the point of release is the critical factor in giving high velocity to the object being projected. Four axes of rotation are often used: Planted foot opposite side of throwing arm Hip opposite throwing arm The spine Shoulder of throwing arm Note: Linear velocity = radius of rotation x angular velocity The radius of rotation is the perpendicular distance between the contact point and the axis of rotation.

36. The open kinetic link model The radius of rotation for throwing extends from the finger tips to the hips. The linear velocity of the finger tips is due to the multiplication of the angular velocity from the inter-segmental kinetic chain. Note: The distal segment will rotate with greater angular velocity because the segment�s rotational inertia is less than the body�s rotational inertia.

37. The open kinetic link model The open kinetic link system for throw-like movements has the following characteristics: 1) The system of links has a base, or fixed end and a free open end. 2) The more massive segments are proximal and the less massive segments are at the free end. 3) An external torque is applied to the base of the segment to initiate the system�s motion and give the entire system angular momentum. Inter- segmental torques are internal to the system and, when applied, change the angular velocities of the individual segments of the system.

38. The internal torques The internal torques are used in the backswing of a throw to connect the segments proximally to distally. The lagging back of the distal segments persists until one or more of the following occurs: 1) The acceleration of the proximal segment decreases or ceases. 2) The elastic or structural limit of a joint�s ROM is reached. 3) The stretch reflex is activated.

39. Throw �like movements performed while in the air Once the body leaves the ground, it will contain a certain amount of angular momentum. In order to perform a throw-like movement the lower limbs must provide a stable base by increasing the rotational inertia to off-set the upper body segments angular velocity.

40. Lever versus wheel-axle rotations Segmental angular movements in the body occur by lever or wheel-axle motions. The angular velocity and the radius of rotation are the two variables of concern. Note: a = T / I (mk2) A smaller radius of gyration will result in a larger acceleration for a given torque. A larger radius of gyration will result in a smaller acceleration for a given torque.

41. Instantaneous velocity Linear velocity at the point of release is the aim of throwing (V = r?). Increased radius increase linear velocity, yet, this increases the radius of gyration, which increases the rotational inertia and thus decreases the accelerating torque. Wheel-axle mechanisms allow for a smaller radius of gyration giving greater angular velocity and hence greater instantaneous linear velocity. Thus, wheel-axle mechanisms are used in throwing.

42. Summary PROJECTILES LAUNCHED HORIZONTALLY Why does the vertical component of velocity for a projectile change with time, whereas the horizontal component of velocity doesn't? The force of gravity on a projectile acts only vertically, hence only the vertical component of a projectile changes with time. There is no component of gravity in the horizontal direction, hence horizontal motion is constant.

43. Summary UPWARDLY LAUNCHED PROJECTILES A rock is thrown upward at an angle. What happens to the horizontal component of its velocity as it rises? As it falls? The horizontal component remains unchanged (in the absence of air drag). A rock is thrown upward at an angle. What happens to the vertical component of its velocity as it rises? As it falls? The vertical component changes the same way that a ball tossed straight upward changes, and changes on the way down just as a rock does in free fall.

44. Web page for practice in projectile calculations http://id.mind.net/~zona/mstm/physics/mechanics/curvedMotion/projectileMotion/commonQuestionsCalculator/commonQuestionsCalculator.html

45. Further questions An object is thrown horizontally off the top of a 45 m high building at a velocity of 20 m/s.� How far from the base of building (in the horizontal direction) will it fall? (ans. 60 m) An object is thrown horizontally off the top of a building at a velocity of 20 m/s and hits the ground 60 m away.� How high up was the building? (ans. 45 m).� You should be able to understand that this is the same problem as above but you are just solving for different things.� An object is thrown from the top of a building and another object is dropped simultaneously.� Which reaches the ground first? (ans. both reach at the same time)