Back to Dimensional Analysis

Exercise 1

Avg difficulty 3.0

Consider a projectile, launched with initial velocity v0v_{0}, at an angle θ\theta. To study its motion we may guess that these are the relevant quantities.

quantity
dimensional
formula
horizontal position xx L1M0T0L^{1}M^{0}T^{0}
vertical position yy L1M0T0L^{1}M^{0}T^{0}
initial speed v0v_{0} L1M0T−1L^{1}M^{0}T^{-1}
angle of launch θ\theta L0M0T0L^{0}M^{0}T^{0}
acceleration due to gravity gg L1M0T−2L^{1}M^{0}T^{-2}
time tt L0M0T1L^{0}M^{0}T^{1}
  • Show that {gt/v0,gx/v02,gy/v02,θ}\{\hskip 1.00006ptgt/v_{0},gx/v_{0}^{2},gy/v_{0}^{2},\theta\hskip 1.00006pt\} is a complete set of dimensionless products. (Hint. One way to go is to find the appropriate free variables in the linear system that arises but there is a shortcut that uses the properties of a basis.)

  • These two equations of motion for projectiles are familiar: x=v0cos⁡(θ)tx=v_{0}\cos(\theta)t and y=v0sin⁡(θ)t−(g/2)t2y=v_{0}\sin(\theta)t-(g/2)t^{2}. Manipulate each to rewrite it as a relationship among the dimensionless products of the prior item.

Hefferon, Linear Algebra, Vector Spaces, Dimensional Analysis, Exercise 1

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Adapted from Linear Algebra © 2016 Jim Hefferon. Licensed under CC BY-SA 3.0 US. Source: Linear Algebra. Imported at revision df2262e.