Science

• Projectile motion is a form of motion where an object moves in a curved path
• Projectile


  The object that is thrown or projected and exhibits projectile motion
• Trajectory


  The curve-like path undertaken by a projectile
• Projectile motion consists of horizontal and vertical motion working independently
• Horizontal motion of a projectile

  • No resistance in any form (air resistance neglected)
  • Constant horizontal velocity (represented as vx)
  • Horizontal acceleration is zero
  • Horizontal distance is called range (represented as dx)
• Vertical motion of a projectile

  • Force acting is gravity (represented as g or ay)
  • Constant acceleration due to gravity is -9.8 m/s^2
  • Vertical velocity is not constant (represented as vy)
  • Vertical distance is called height (represented as dy)
• Solving a problem on horizontal launched projectile

  1. Given: Horizontal distance (dx), Initial velocity (vi), Acceleration due to gravity (g)

  2. Find: Height of table (dy), Final velocity (vf)

  3. Step 1: Find time using dx and vi

  4. Step 2: Use time to find dy

  5. Step 3: Use dy to find vf
• The negative sign in the final velocity indicates the downward direction
• Kinematic equations at constant speed

  d = vt (displacement = velocity x time)
• Kinematic equations under constant acceleration


  • d = (v_initial + v_final)/2 * t (displacement = average velocity x time)
  • v_final = v_initial + a*t (final velocity = initial velocity + acceleration x time)
  • v_final^2 = v_initial^2 + 2*a*d (final velocity^2 = initial velocity^2 + 2*acceleration*displacement)
  • d = v_initial*t + 1/2*a*t^2 (displacement = initial velocity*time + 1/2*acceleration*time^2)
• Displacement can be along the x-axis or y-axis
• Distance and displacement are the same if the object moves in one direction without changing direction
• Projectile


  An object only under the influence of gravity, with air resistance ignored
• Horizontal velocity (v_x) remains constant for a projectile
• Vertical velocity (v_y) changes by the acceleration due to gravity (-10 m/s^2) each second for a projectile
• Speed


  The magnitude of velocity, always positive
• Velocity


  A vector quantity with both magnitude and direction
• For a projectile, velocity decreases as it moves upwards, but speed decreases on the upward path and increases on the downward path
• Types of projectile trajectories

  • Ball rolling off a cliff
  • Ball kicked off the ground
  • Ball kicked off a cliff/building
• Equations for ball rolling off a cliff

  • h = 1/2 * a_y * t^2 (height = 1/2 * acceleration_y * time^2)
  • r = v_x * t (range = horizontal velocity * time)
• Equations for ball kicked off the ground

  • t_max = v_y_initial / g (time to maximum height = initial vertical velocity / gravity)
  • h_max = v^2 * sin^2(theta) / (2*g) (maximum height = velocity^2 * sin^2(angle) / (2*gravity))
  • r = v^2 * sin(2*theta) / g (range = velocity^2 * sin(2*angle) / gravity)
• The speed at the highest point of a projectile's trajectory is the same as the speed when it lands
• Equation for ball kicked off a cliff/building

  y_final = y_initial + v_y_initial*t + 1/2*a_y*t^2 (final vertical position = initial vertical position + initial vertical velocity*time + 1/2*acceleration_y*time^2)
• If objects are at the same height, the speed will be the same, so the final speed at point C is the same as the initial speed at point A. The vertical velocity is positive at point A but negative at point C, but the magnitudes are the same.
• Trajectory of a ball kicked off a cliff/building at an angle

  1. Goes up

  2. Goes back down
• Equation: y_final = y_initial + v_y_initial*t + 1/2*a_y*t^2

  y_final is the final vertical position, y_initial is the initial vertical position, v_y_initial is the initial vertical velocity, a_y is the vertical acceleration (negative 9.8 m/s^2), t is the time
• v_y


  Vertical velocity = v*sin(theta)
• Finding time to hit the ground using quadratic formula

  1. t = (-b +/- sqrt(b^2 - 4ac)) / (2a)

  2. Where a = 1/2*a_y, b = v_y_initial, c = y_initial
• Alternative method to find time

  1. Find time from A to B

  2. Find time from B to C

  3. Add the two times
• Equation: t = v_y_initial / g


  Time from A to B
• Equation: h + y_initial = 1/2*a_y*t^2

  Time from B to C
• Equation: v_final = sqrt(v_x^2 + v_y^2)


  Calculate final speed just before hitting the ground
• Equation: theta = arctan(v_y/v_x)

  Calculate final angle just before hitting the ground
• Problem 1: Ball rolls horizontally off a 200m high cliff at 20 m/s


  1. Find time to hit ground

  2. Find horizontal distance traveled
• Equation: h = 1/2*a*t^2

  To find height of cliff
• Equation: range = v_x*t


  To find horizontal distance traveled
• Problem 2: Ball released from rest at 800m height


  Find time to hit ground
• Problem 3: Ball thrown straight down at 30 m/s from 800m height


  Find time to hit ground
• Equation: d_y = v_y_initial*t + 1/2*a_y*t^2


  Used to solve quadratic equation for time