Prerequisite materialVelocity and Position Can Be Derived from Acceleration

Motion in a Plane and Parabolic Trajectories

One ball is thrown horizontally while another is released from rest at the same height. Their motion is resolved into horizontal and vertical components to determine the time each takes to reach the floor.

高さ4.9 mから一方を横に投げ、他方を同時に放す。先に床に着くのは?

The equations explain the phenomenon

  1. 1Position is represented by an arrowr=⁠(x, y)……
  2. 2Velocity is obtained from the change in positionv=⁠drdt=⁠(dxdt, dydt)……
  3. 3Acceleration is obtained from the change in velocitya=⁠dvdt=⁠(dvxdt, dvydt)……
  4. 4The equation of motion is resolved into horizontal and vertical componentsm dvxdt = 0⁠,m dvydt = −mg……
  5. 5The velocity of a ball thrown horizontally is calculatedvx = v₀⁠,vy = −g t……
  6. 6The position of a ball thrown horizontally is calculatedx = v₀ t⁠,y = h − ½ g t²……
  7. 7The landing time for a horizontal launch is calculatedt=⁠√(2hg)……
  8. The next sections examine motion with air resistance and launches at an angle.
  9. 8Air resistance is includedm dvdt=⁠m (0, −g)−⁠k |v| v……
  10. 9Air resistance is resolved into horizontal and vertical componentsm dvydt=⁠−m g−⁠k √(vx² + vy²) vy……
  11. 10The position of a ball launched at an angle is calculated without air resistancex = v₀ cosθ t⁠,y = v₀ sinθ t − ½ g t²……
  12. 11Time t is eliminated to obtain the trajectory equationy=⁠tanθ x−⁠g x²2 v₀² cos²θ……
  13. 12The horizontal range on returning to the launch height is calculatedR=⁠v₀² sin2θg……
  14. 13The angle for maximum range with air resistance is calculatedθmaximum range≒⁠39.0°<⁠45°……