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Kinematic Equations and Projectile Types

Jun 20, 2025

Overview

This lecture reviews essential kinematic equations and explains how to solve common projectile motion problems involving different types of trajectories.

Basic Kinematic Equations

  • Displacement for constant velocity: ( d = vt )
  • Final velocity with constant acceleration: ( v_f = v_i + at )
  • Displacement using average velocity: ( d = \frac{v_i + v_f}{2}t )
  • Displacement using initial velocity and acceleration: ( d = v_it + \frac{1}{2}at^2 )
  • Velocity-squared equation: ( v_f^2 = v_i^2 + 2ad )
  • Displacement (( d )) means the difference between final and initial positions, can apply in ( x ) or ( y ) directions.

Horizontal Launch from a Height (Type 1)

  • Height: ( h = \frac{1}{2}gt^2 ) (uses vertical motion only, ( v_{iy} = 0 ))
  • Range: ( R = v_{x}t ) (( v_x ) is constant; ( a_x = 0 ))
  • To find speed before impact: ( v = \sqrt{v_x^2 + v_y^2} ) where ( v_{yf} = gt )
  • Angle before hitting ground: ( \theta = \arctan \left( \frac{v_y}{v_x} \right) )_

Angled Launch from Ground (Type 2)

  • Initial velocity components: ( v_x = v \cos \theta ), ( v_y = v \sin \theta )
  • Time to max height: ( t_{up} = \frac{v \sin \theta}{g} )
  • Total flight time: ( t_{total} = \frac{2v \sin \theta}{g} )
  • Maximum height: ( h_{max} = \frac{v^2 \sin^2 \theta}{2g} )
  • Range: ( R = \frac{v^2 \sin 2\theta}{g} )
  • At landing, speed and angle match launch values (symmetrical path)._

Angled Launch from a Height (Type 3)

  • Use: ( y_f = y_i + v_{iy}t + \frac{1}{2} a t^2 ) (for vertical position)
  • Set ( y_f = 0 ) to solve for time ( t ) (may require quadratic formula).
  • Alternative: Find time up then time down, sum for total time.
  • Range: ( R = v_x t_{total} ) where ( v_x = v \cos \theta )
  • Do not use the symmetrical range formula unless the trajectory is symmetrical.

Calculating Final Speed and Angle Upon Impact

  • ( v_x ) is constant throughout the motion.
  • Final vertical velocity: ( v_{yf} = v_{iy} - gt )
  • Impact speed: ( v = \sqrt{v_x^2 + v_{yf}^2} )
  • Impact angle: ( \theta = \arctan \left( \frac{|v_{yf}|}{v_x} \right) )
  • Describe angle as degrees below the horizontal or relative to the positive ( x )-axis as appropriate.

Key Terms & Definitions

  • Displacement — The straight-line distance and direction from initial to final position.
  • Range (( R )) — Horizontal distance traveled by the projectile.
  • Maximum Height (( h_{max} )) — Maximum vertical displacement above launch point.
  • Projectile Motion — Motion involving both horizontal and vertical components under constant acceleration due to gravity._

Action Items / Next Steps

  • Memorize the core kinematic and projectile motion equations.
  • Practice distinguishing which equation to use for each type of trajectory.
  • Complete assigned projectile motion practice problems, focusing on separating ( x ) and ( y ) components.