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Projectile Motion

Projectile Motion. A branch of kinematics that deals with the motion of objects travelling through the air. Chapter 1 in the text. Assumptions and Facts. For many cases, the force of air resistance on a ballistic can be considered negligible. Shot-put, jumping, basketball shot

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Projectile Motion

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  1. Projectile Motion A branch of kinematics that deals with the motion of objects travelling through the air. Chapter 1 in the text Dr. Sasho MacKenzie - HK 376

  2. Assumptions and Facts • For many cases, the force of air resistance on a ballistic can be considered negligible. • Shot-put, jumping, basketball shot • If free in the air, an object has known vertical (-9.81 m/s2) and horizontal (0 m/s2) accelerations. • Vertical and horizontal motion are independent. • Fired versus dropped bullet • The above allow a special set of rules (equations) to analyze projectile motion Dr. Sasho MacKenzie - HK 376

  3. Biathlon Example • We are going to use the example of a bullet fired from a biathlete’s rifle to garner a deeper understanding of projectile motion. • The horizontal and vertical motion of the bullet will be considered independently. • Using graphical techniques, geometry, and algebra we will arrive at equations describing the position of the bullet as a function of time. Dr. Sasho MacKenzie - HK 376

  4. Missed Target • Using an Anschutz rifle, which has a muzzle velocity of 340 m/s, a Norwegian Biathlete missed the target. The rifle was being held at shoulder height (1.6 m) and parallel to the ground when fired. • The bullet takes 0.57 s to fall to the ground after leaving the muzzle [this time is usually not provided]. • How far does the bullet travel horizontally? Dr. Sasho MacKenzie - HK 376

  5. Bullet’s Path (not to scale) 1.6 m Dr. Sasho MacKenzie - HK 376

  6. y x Horizontal Velocity (Vx) 340 m/s Vx 0.57 Time (s) Dr. Sasho MacKenzie - HK 376

  7. y x Horizontal Displacement (Dx) 194 m Dx 0.57 Time (s) Dr. Sasho MacKenzie - HK 376

  8. Look at Bigger Picture • At this point there are no NEW concepts, we already know that D = Vt. • However, we were given the time (t = 0.57) it took the bullet to fall to the ground. • Typically, this information will not be provided and must be calculated by you! • The motion of the bullet in the vertical direction must be analyzed to determine how long a projectile spends in the air. Dr. Sasho MacKenzie - HK 376

  9. y x Vertical Acceleration (ay = g) ay 0 Time (s) 0.57 -9.81 Dr. Sasho MacKenzie - HK 376

  10. y x Vertical Velocity (Vy) Vy 0 Time (s) 0.57 -5.6 m/s Dr. Sasho MacKenzie - HK 376

  11. y x Vertical Displacement (Dy) 1.6 Dy 0 Time (s) 0.57 Closer look on next slide! Dr. Sasho MacKenzie - HK 376

  12. Displacement is the area under the velocity curve Vy t 0 This area is a triangle with Base = t, and Height = Vf = ayt Vf The area of a triangle is ½ Height x Base Therefore, displacement = ½ ayt2 If there is an initial velocity, then you must add on the rectangular area determined by: Vinitialt Dr. Sasho MacKenzie - HK 376

  13. Vyinitialt ½ayt2 Displacement if Vyinitial is NOT zero t 0 Vy There are now 2 areas to add together. Vi Vf Dr. Sasho MacKenzie - HK 376

  14. Tips and Equation Rearrangements • If the initial vertical velocity is zero, then • If the object’s initial vertical position = the final vertical position, then Dr. Sasho MacKenzie - HK 376

  15. Shot Put Example • A shot put is released from a height of 2 m with a velocity of 15 m/s at an angle of 39 above the horizontal. • How long does the shot stay in the air? • What is the maximum height of the shot above the ground? • How far does the shot travel horizontally (distance of throw)? 15 m/s 39 2 m Dr. Sasho MacKenzie - HK 376

  16. 1. Find Vxinitial and Vyinitial 15 m/s Vy = sin(39)(15) = 9.4 m/s 39 Vx = cos(39)(15) = 11.7 m/s 9.4 m/s 11.7 m/s Dr. Sasho MacKenzie - HK 376

  17. At the top, the shot will have no vertical velocity 9.4 m/s 2 m Finding Time to Peak Height Vy = ayt, therefore; t = Vy / ay t = 9.4/ 9.81 = 0.96 s It takes 0.96 s for the shot to reach its peak height Finding Peak Height 2. Analyze Up and Down Separately UP Therefore, V = 9.4 m/s Dypeak = Dyinitial + Vyinitialt + ½at2 = 2 + (9.4)(.96) + ½ (-9.81)(.96)2 = 2 + 9.024 - 4.52 = 2 + 4.5 = 6.5 m above the ground 4.5 m Dr. Sasho MacKenzie - HK 376

  18. At the top, the shot will have no vertical velocity. From analyzing the “UP” portion of the flight, we know the shot is 6.5 m above the ground at the top. 6.5 m Finding Total Flight Time Finding Time to Fall (Vyinitial= 0) Timetotal = TimeUp + TimeDown = 0.96 + 1.15 = 2.11 s Finding Throw Distance Dxfinal = Dxinitial + Vxinitialt = 0 + (11.7)(2.11) = 0 + 24.7 = 24.7 m from point of release 2. Analyze Up and Down Separately DOWN t = 1.15 time to fall from peak height of 6.5 m Dr. Sasho MacKenzie - HK 376

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