Rectilinear motion definitions; projectile at 244 m/s

Uniformly rectilinear motion

A particle moves along a straight line with constant velocity, so its acceleration is zero: v = constant, x = x₀ + vt. The x–t graph is a straight line and the v–t graph is horizontal.

Uniformly accelerated rectilinear motion

A particle moves along a straight line with constant acceleration a. Then:

• v = u + at
• s = ut + ½at²
• v² = u² + 2as where u is the initial velocity, v the velocity after time t and s the displacement. The v–t graph is a straight inclined line and the x–t graph is a parabola. Free fall (a = g = 9.81 m/s²) is the common example.

Projectile problem

Take A as origin. Initial speed v₀ = 244 m/s, target at x = 3658 m, y = 610 m. The equation of the trajectory (no air resistance):

x = (v₀ cos α) t
y = (v₀ sin α) t − ½ g t²
y = x tan α − g x² / (2 v₀² cos²α)
  = x tan α − [g x² / (2 v₀²)] (1 + tan²α)

• g x²/(2v₀²) = 9.81 × 3658² / (2 × 244²) = 1102.42
• Substituting: 610 = 3658 tan α − 1102.42(1 + tan²α)
• 1102.42 tan²α − 3658 tan α + 1712.42 = 0
• tan α = [3658 ± √(3658² − 4(1102.42)(1712.42))] / (2 × 1102.42) = (3658 ± 2414.5)/2204.84

So tan α = 0.5640 or 2.7542.

Answer: α = 29.4° (low trajectory) or α = 70.0° (high trajectory). Both angles hit B; the firing angle normally used is 29.4°.