MODULE 1

Cards (24)

  • The purpose of this module is to help you understand the horizontal and vertical motion of a projectile and the relationship between the angle of release and the height and the range of the projectile
  • This module contains one lesson: Projectile Motion
  • After going through this module, you are expected to
    • Describe the horizontal and vertical motion of a projectile
    • Investigate the relationship between the angle of release and the height and range of the projectile
  • Projectile
    An object thrown with an initial velocity and acted upon by the earth's pull of gravity
  • Trajectory
    The curved path taken by the projectile
  • Range
    The horizontal distance traveled by a projectile within the total time of its flight
  • Projectile motion

    • Combination of vertical and horizontal motion which are independent of each other
    • Horizontal motion is constant while vertical motion is changing
  • Equations of projectile motion
    1. Vertical motion: h=gt^2/2
    2. Horizontal motion: R=vxt
  • Projectile thrown at an angle
    Horizontal velocity remains constant, vertical velocity changes
  • Maximum height

    When vertical velocity is zero
  • Projectile motion at complementary angles have the same range
  • Sample problem: Projectile thrown at 20 degrees

    Given: Vi=50m/s, θ=20°
    Formula: R=vi^2sin2θ/2g
    Solution: R=164m
  • Height of the projectile

    h=(visinθ)(t) - (1/2)gt^2
  • Range R
    R= vi^2sin 2θ/g
  • The range R is calculated using this equation and is applicable to all values of θ
  • An object that is thrown at 30° into the air will have the same range when it is thrown at 60°
  • The object when thrown at 70°, 20°, and 10° will have the same ranges as it was thrown in 20°, 70°, and 80° respectively
  • The maximum range obtained is 45°
  • Sample Problem: What must be the horizontal distance of the ball if it is thrown with initial velocity of 50.0 m/s at an angle of 20.0°?
    1. Given: Vi = 50.0m/s, θ = 20.0°, Unknown: R = ?
    2. Formula: R= vi^2sin2θ/2g
    3. Solution: R= (50m/s)^2sin2(20°)/(9.8m/s^2)
    4. Answer: R = 164 m
  • Projectile Motion and Complementary Angles
  • When an object is thrown at 15°, 30°
    It will have the same range
  • The launch angle that will have the same range as a projectile launched at 20°
    70°
  • In the absence of air resistance, the angle to get the maximum range is 45°
  • In the Javelin throw event, the angle to attain the farthest distance is 45°