projectile motion

Definition of Angle

  • An angle in physics is defined based on its orientation concerning a reference direction.
  • Specifically, the positive direction of the x-axis is commonly used as a reference point for measuring angles counterclockwise to a given vector.
  • If the angle given does not conform to this standard definition, the cosine function may not accurately provide the x-component of the vector.

Understanding Components of a Vector

  • X-Component: Determined from the cosine of the angle when measuring from the reference direction.
  • Y-Component: Determined from the sine of the angle.
  • Caution is advised when interpreting these shortcuts based on the geometric orientation of the problem, as real-life scenarios might involve different angles and orientations.

Example of Projectile Motion

  • Initial Conditions:
    • A ball is thrown at an angle of 30 degrees above the horizon with an initial velocity of 28 m/s.
    • The release height is given as 1.8 meters above the ground.

Problem 1: Horizontal Distance Travelled

  • Goal: Calculate how far the ball travels horizontally before hitting the ground.
  • The following questions are considered:
    • How far does the ball travel?
    • How fast is it going when it hits the ground?
    • At what angle is the ball traveling when it hits the ground?

Kinematic Equations in Projectile Motion

  • X-Direction:
    • In the x-direction, there is no acceleration (uniform motion).
    • The equation used is:
      x=x0+viximesexttimex = x_0 + v_{i_x} imes ext{time}
  • Y-Direction:
    • The y-direction experiences gravitational acceleration, denoted by −9.8extm/s2-9.8 ext{ m/s}^2.
    • The initial height is given as y0=1.8extmy_0 = 1.8 ext{ m}.
    • The initial velocity in the y-direction can be calculated using:
      viy=viimesextsin(heta)=28imesextsin(30)=14extm/s.v_{i_y} = v_i imes ext{sin}( heta) = 28 imes ext{sin}(30) = 14 ext{ m/s}.
    • The y-position can be modeled by the equation:
      y = y_0 + v_{i_y} imes ext{time} + rac{1}{2} a_{y} ( ext{time})^2.

Solving for Time in the Air

  • Final y-position when the ball hits the ground equals zero:
    0 = 1.8 + 14 imes ext{time} - rac{1}{2} imes 9.8 imes ( ext{time})^2.
  • This leads to solving a quadratic equation in the form 0=−4.9(exttime)2+14(exttime)+1.80 = -4.9 ( ext{time})^2 + 14 ( ext{time}) + 1.8.
  • The roots are found using the quadratic formula: ext{time} = rac{-b ext{±} ext{sqrt}(b^2 - 4ac)}{2a}, where:
    • a=−4.9,b=14,a = -4.9, b = 14, and c=1.8.c = 1.8.

Solution for Time

  • Solving gives two potential solutions for time; one will be discarded as it is negative, leaving the positive time: exttime=2.98exts.ext{time} = 2.98 ext{ s}.

Find Distance in X Direction

  • After determining the time in the air, the distance in the x-direction can now be calculated using the previously established equation:
    • Initial velocity in x-direction:
      vix=28imesextcos(30)=24.25extm/s.v_{i_x} = 28 imes ext{cos}(30) = 24.25 ext{ m/s}.
    • Final x-distance calculated:
      x=0+24.25imes2.98=72extm.x = 0 + 24.25 imes 2.98 = 72 ext{ m}.

Check Results with Reality

  • Verify outcomes against practical expectations (e.g., throwing distance not exceeding physical limits).

Final Velocity When Hitting the Ground

  • To understand the final velocity's components:
    • Summarizing the velocities at impact:
    • X Component: Remains constant at 24.25 m/s due to no acceleration.
    • Y Component: Calculated using
      vfy=viy+ayimesexttime=14+(−9.8)imes2.98=−15.2extm/s.v_{f_y} = v_{i_y} + a_y imes ext{time} = 14 + (-9.8) imes 2.98 = -15.2 ext{ m/s}.

Magnitude of Resultant Velocity

  • Use the Pythagorean theorem to find the resultant velocity:
    vf=extsqrt((vfx)2+(vfy)2)=extsqrt((24.25)2+(−15.2)2)=28.6extm/s.v_f = ext{sqrt}((v_{f_x})^2 + (v_{f_y})^2) = ext{sqrt}((24.25)^2 + (-15.2)^2) = 28.6 ext{ m/s}.

Angle of Impact

  • To find the angle at impact (C6), use the tangent function: an( heta) = rac{v_{f_y}}{v_{f_x}} ightarrow ext{angle} = an^{-1}igg( rac{-15.2}{24.25}igg).
    • This angle gives the trajectory concerning the horizontal.

Adjusted Examples

  • Hit a Wall Before Ground:
    • New case where trajectory hits a wall (D1.5 m high wall at x=15 m):
    • The time of travel may now rely on the x-direction rather than y-direction.
    • Important: Understand that time of flight can vary based on the new endpoint (the wall).

Final Questions to Consider

  • What is the angle of the velocity vector when returning to the same height from which it was thrown?
    • It's crucial to distinguish between the angle of release and the angle of return, considering factors such as height and horizontal distance travelled.
  • Assure to consider all possible cases while applying projectile motion equations systematically, ensuring correct application of both component analyses and time calculations.
  • Recognize that in different scenarios (such as hitting a wall), one must adjust calculations to fit the physical situation rather than adhering rigidly to a predefined method.

Summary

  • Projectile motion problems require a detailed and systematic approach to resolving both the horizontal and vertical components while acknowledging that motion characteristics may differ based on varying conditions (e.g., initial height, target distance).