Projectile Motion

• Projectile


  An object upon which the only force acting on it is gravity. In Physics, this is called projectile motion.
• Trajectory


  The path that the projectile follows
• Projectile motion


  • Deals with the initial component of velocity (vy)
  • Has both vertical and horizontal components of velocity
• Solving projectile motion


  1. Horizontal motion: vx = v cos θ, ax = 0, vfx = vix, dx = vixt

  2. Vertical motion: vy = v sin θ, ay = -g, vfy = viy + gt, dy = viyt + 1/2 gt^2
• The acceleration due to gravity (ag) is ALWAYS negative (-). Any object affected only by gravity (a projectile or an object in free fall) has an acceleration of -9.8 m/s2, regardless of the direction.
• Projectile travels along the trajectory with a constant horizontal velocity (vx) and a changing vertical velocity (vy). The vertical velocity changes by a magnitude of 9.8 m/s each second.
• Solving projectile motion problems

  1. Calculate initial vertical velocity (viy = v sin θ)

  2. Calculate time to reach maximum height (vfy = 0, t = -viy/g)

  3. Calculate maximum height (dy = viyt + 1/2 gt^2)

  4. Calculate total time of flight (t = 2(-viy/g))

  5. Calculate horizontal displacement (dx = vixt)
• A projectile is an object that has the following characteristics: 1) The only force acting on it is gravity, 2) It has both horizontal and vertical motion
• Finding HorVert


  1. Use the range and maximum height equation to calculate the horizontal distance a baseball

  2. If the initial velocity is 55 m/s and the ball is thrown at an angle of: 20°

  3. 40°

  4. 60°

  5. 70°
• The trajectory diagram shows the position of the ball after each consecutive second
• Free-falling Inlove with You!

  1. The only force acting on it is a gravity

  2. The acceleration is directed downwards and has a value of -9.8 m/s2

  3. Once projected, it continues its horizontal motion without any need of a force

  4. As it rises, its vertical velocity (vy) decreases; as it falls, its velocity (vy) increases

  5. As it travels through the air, its horizontal velocity remains constant

  6. For 0° ≤ θ < 45°, the range (R) increases with an increasing angle of projection

  7. For 45° < θ ≤ 90°, the range (R) decreases with an increasing angle of projection

  8. Range is maximum when θ = 45°
• What is the Best Angle for Launching a Catapult?

  1. Use rubber bands to create potential energy through tension or torsion

  2. Use the spoon as the launcher

  3. Use the protractor to get the angle of release of a projectile

  4. Test the design and see which angle allows the projectile to travel the furthest

  5. Start with 30 degrees for the first launch

  6. Repeat the process at 45 degrees and 60 degrees
• When a projectile is launched at a certain angle above the ground, its range is calculated using the equation:

  
• For a given initial velocity Vi, dx is maximum when sin^2 θ =1. Thus, θ = 45. The maximum height can be solved using the equation:

  
• Study the following picture