Projectile Motion A branch of kinematics that deals with the motion of objects travelling through the air. Chapter 1 in the text 2/17/2019 Dr. Sasho MacKenzie - HK 376

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 These assumptions and facts 2/17/2019 Dr. Sasho MacKenzie - HK 376

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. 2/17/2019 Dr. Sasho MacKenzie - HK 376

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? 2/17/2019 Dr. Sasho MacKenzie - HK 376

Bullet’s Path (not to scale) 1.6 m 2/17/2019 Dr. Sasho MacKenzie - HK 376

Horizontal Velocity (Vx) 340 m/s Vx Time (s) 0.57 y x 2/17/2019 Dr. Sasho MacKenzie - HK 376

Horizontal Displacement (Dx) Time (s) 0.57 y x 2/17/2019 Dr. Sasho MacKenzie - HK 376

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. 2/17/2019 Dr. Sasho MacKenzie - HK 376

Vertical Acceleration (ay = g) Time (s) 0.57 -9.81 y x 2/17/2019 Dr. Sasho MacKenzie - HK 376

Vertical Velocity (Vy) Time (s) 0.57 -5.6 m/s y x 2/17/2019 Dr. Sasho MacKenzie - HK 376

Vertical Displacement (Dy) 1.6 Dy Time (s) 0.57 Closer look on next slide! y x 2/17/2019 Dr. Sasho MacKenzie - HK 376

Displacement is the area under the velocity curve Vy t 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 2/17/2019 Dr. Sasho MacKenzie - HK 376

Displacement if Vyinitial is NOT zero Vy There are now 2 areas to add together. Vyinitialt Vi ½ayt2 Vf 2/17/2019 Dr. Sasho MacKenzie - HK 376

Tips and Equation Rearrangements If the initial vertical velocity is zero, then If the object’s initial vertical position = the final vertical position, then 2/17/2019 Dr. Sasho MacKenzie - HK 376

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 2/17/2019 Dr. Sasho MacKenzie - HK 376

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 2/17/2019 Dr. Sasho MacKenzie - HK 376

2. Analyze Up and Down Separately 9.4 m/s At the top, the shot will have no vertical velocity 2 m UP Therefore, V = -9.4 m/s Vy = ayt, therefore; t = Vy / ay Finding Time to Peak Height t = -9.4 / -9.81 = 0.96 s It takes 0.96 s for the shot to reach its peak height Finding Peak Height 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 2/17/2019 Dr. Sasho MacKenzie - HK 376

2. Analyze Up and Down Separately 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 DOWN Finding Time to Fall (Vyinitial= 0) Timetotal = TimeUp + TimeDown = 0.96 + 1.15 = 2.11 s Finding Total Flight Time Finding Throw Distance Dxfinal = Dxinitial + Vxinitial t = 0 + (11.7)(2.11) = 0 + 24.7 = 24.7 m from point of release t = 1.15 time to fall from peak height of 6.5 m 2/17/2019 Dr. Sasho MacKenzie - HK 376