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Projectile Motion

This chapter explores the principles of projectile motion, focusing on the motion of objects traveling through the air. Examples and equations are utilized to analyze the motion of projectiles in both the vertical and horizontal directions.

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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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