Study Notes on Projectile Motion

Projectile Motion

  • Definition: A motion of an object is termed as projectile motion, characterized by the object's motion along a curved path under the influence of gravity.

Key Concepts of Projectile Motion

  • The motion can be broken down into vertical and horizontal components.
    • Objects in projectile motion typically start from the same height and end up at the same height.
Changes in Velocity
  • Vertical Motion:

    • As an object moves upward, its velocity decreases until it reaches maximum height, where the velocity is zero.
    • After reaching maximum height, the object begins to fall, causing its velocity to increase again.
  • Acceleration:

    • Regardless of the changes in velocity, acceleration remains constant at 9.8m/s29.8 \, \text{m/s}^2 throughout the motion.
    • This constant acceleration reflects the influence of gravity on the object.

Types of Projectile Motion

  1. Horizontal Projectile Motion:

    • The object is projected horizontally, following a parabolic path as it moves downward under gravity.
    • Example: A person throws a ball horizontally.
    • The initial velocity is horizontal; therefore, it is denoted as vixv_{ix} or vxv \cdot x.
    • Gravity acts downward, causing a change in vertical velocity while maintaining constant horizontal velocity.
  2. Vertical Projectile Motion:

    • Involves upward and downward motion where initial vertical velocity may vary.
    • Consideration of maximum height and gravitational acceleration.
Analysis of Projectile Motion
  • Velocity as a Vector Quantity:
    • Velocity includes both magnitude (speed) and direction.
    • When analyzing projectile motion:
    • The horizontal velocity vxv_{x} remains constant; thus, extaccelerationinthexdirection=0ext{acceleration in the x-direction} = 0.
      • This is because acceleration is defined as the change in velocity over time, which is zero when velocity is constant.
    • The vertical velocity vyv_{y} increases as the object rises and decreases as it falls due to gravitational acceleration.
      • This increase represents the constant acceleration of 9.8m/s29.8 \, \text{m/s}^2 in the y-direction.
Reference Points in Projectile Motion
  • The reference for height (y-coordinate) can be defined as either the maximum height (top of the throw) or the ground. Both references are valid as they are arbitrary.
  • For calculations, the choice of downward as negative or positive impacts the signs of your results, especially in determining acceleration due to gravity.
    • For downward direction:
    • ay=9.8m/s2a_y = -9.8 \, \text{m/s}^2
    • For upward direction:
    • ay=9.8m/s2a_y = 9.8 \, \text{m/s}^2
Kinematic Equations and Projectile Motion
  • Since projectile motion involves constant acceleration, standard kinematic equations apply:
    • v<em>y=v</em>iy+aytv<em>y = v</em>{iy} + a_y t
    • y=v<em>iyt+12a</em>yt2y = v<em>{iy} t + \frac{1}{2} a</em>y t^2
    • v<em>y2=v</em>iy2+2a<em>y(yy</em>0)v<em>y^2 = v</em>{iy}^2 + 2a<em>y(y - y</em>0)
  • The position and motion in the x-direction can be described by a separate set of equations due to the lack of acceleration in that dimension.

Problem Solving in Projectile Motion

  • Methodology for Solving Problems:
    • Read the problem several times to understand.
    • Identify and write down known values for x and y axes in a chart or diagram.
    • Use the relevant kinematic equations as required by the specific problem.
Example Problems
  1. Horizontal Projectile (Building Example):

    • Known Values:
      • Initial horizontal velocity vix=12.2m/sv_{ix} = 12.2 \,\text{m/s}
      • Height calculation gives extheight=14.8metersext{height} = 14.8 \, \text{meters}
  2. Vertical Motion (Jump Example):

    • Given information regarding time airborne yields calculations for how high an individual jumps (e.g. 4.05meters4.05 \, \text{meters} max height) without dividing time by two if asked for total airborne duration.
Advanced Problem Solving Concepts
  • In more complex examples involving angles, use trigonometric functions to calculate resultant velocities and angles.
  • Example: To find the net velocity where height and horizontal motion are combined, utilize:
    • V=v<em>x2+v</em>y2V = \sqrt{v<em>x^2 + v</em>y^2} and
    • Use tanθ=V<em>yV</em>x\tan\theta = \frac{V<em>y}{V</em>x} to determine the angle in reference to the horizontal.
  • When determining angles, clarify whether they are above or below the reference axis.

Conclusion

  • Understanding projectile motion involves dissecting the components of motion, maintaining awareness of gravitational effects, applying kinematic equations, and accurately solving problems with respect to the frame of reference in which they are set.