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Projectile motion

Projectile motion. ballistic motion free fall. Assumptions. The Earth’s surface is flat (a plane) Gravitational acceleration g = 9.81 m/s 2 There is no air resistance ( Later on during the course, we will return to the subject with more general assumptions ).

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Projectile motion

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  1. Projectile motion ballistic motion free fall

  2. Assumptions • The Earth’s surface is flat (a plane) • Gravitational acceleration g = 9.81 m/s2 • There is no air resistance ( Later on during the course, we will return to the subject with more general assumptions )

  3. Example1: A stone is thrown directly upwards with an initial velocity of 40 m/s. • How long does it stay in the air ? • What is the highest point of the trajectory? Approximate solution without a calculator: Answers: Flight time is 8 sec. Highest point 80 m

  4. Example2: A bullet is shot with an initial speed of 400 m/s in 35 deg angle. • How long does it stay in the air ? • b) What is the highest point of the trajectory? • c) How far does the bullet fly horizontally ? The bullet experiences 2 different types of motions . Horizontally it is in a uniform motion with a constant speed. Vertically its motion is uniformly accelerating with a = - 9.81 m/s2. 1) Divide the initial velocity vector into horizontal and vertical components: v0 sinα=229.4 v0=400 m/s v0 cosα=400*cos35o =327.7 m/s

  5. 2. Calculate the flight time. Vertical velocity is initially 229.4 m/s. It changes with -9.81 every second. => After about 23 seconds velocity is zero and the bullet is at the ceiling point. After about 46 seconds the bullet hits the ground. Mathematically: The equation for vertical velocity is: vy = v0 sinα – g t Substituting vy = 0 we get the time when the bullet is at highest. t = v0 sinα / g ( here 23.4 s) 3. Calculate the highest point Vertical velocity is initially 229.4 m/s. At the ceiling point it is 0. The average velocity during the rise is 229.4/2. The highest y- coordinate is 229.4/2 m/s* 23.4 s = 2684 m = 2. 6 km. Mathematically: The highest point ymax = v0 sinα / 2* v0 sinα / g => ymax = v02 (sinα)2 / (2g)

  6. 3. Calculate the horizontal distance of the flight Horizontal velocity is initially and all the time 327.7 m/s. Once we know the flight time 46.8 s , it is easy to calculate the distance x = 327.7 m/s * 46.8 s = 15336 m = 15.3 km • Mathematically: The horizontal velocity is v0 cosα • and the flight time = 2 v0 sinα / g • x = 2 v02 sin α cos α / g x = v02/g sin(2α) Observations: The horizontal distance is proportional to the square of the initial velocity. sin(2α) = 1 when α = 45o. The shot or throw is longest, when the angle is 45 degrees. ( This is true only when the starting and landing levels are same)

  7. The equations for (x,y) In Olympic games a sportsman throws hammer, which starts its flight at 1.8 m height at the speed of 27.0 m/s. The angle is 47 degrees. Calculate the result: The flight time: Result:

  8. With Mathematica

  9. Problem: Olympic winner of shot put threw 21.32. Assume that the angle was 46 degrees and the ball started at 1.8 m height. Calculate the initial velocity of the throw. Result: The initial velocity v0 was 13.91 m/s and the time = 2.206 s

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