Chapter 6 - Projectiles - Projectile Motion Formulae

7問 • 2年前

Oluwole Akande

通報

問題一覧

Time of Flight =

2Usinθ / g

Time to Reach Greatest Height =

Usinθ / g

Range on Horizontal Plane =

U²sin2θ / g

Equation of Trajectory (Instantaneous Vertical Height) =

y = xtanθ - (gx²(1 + tan²θ) / 2U²)

How is the Equation for Time of Flight Derived

1 - Resolve for Vertical Motion, where u = Usinθ, a = -g, s = 0, and t is unsolved, 2 - Substitute into "s = ut + ½at²", 3 - 0 = T(Usinθ) - ½gT², 4 - 0 = T(Usinθ - gT/2), 5 - We know that either T = 0 or Usinθ - gT/2 = 0, 6 - Rearrange to get T = 2Usinθ / g

How is the Equation for Horizontal Range Derived

1 - Resolve Horizontally, where v = Ucosθ, s = R, and t = T, 2 - Substitute into "s = vt", 3 - R = Ucosθ × T, 4 - We know that T = 2Usinθ / g, therefore R = Ucosθ × 2Usinθ / g, 5 - R = 2U²sinθcosθ / g, 6 - Using the Double Angle Formula, 2sinθcosθ = sin2θ, 7 - Therefore R = U²sin2θ / g

How is the Equation for the Trajectory of a particle Derived

1 - We know that U(x) = Ucosθ, and U(y) = Usinθ, 2 - We know that for horizontal motion, x = vt = Ucosθ × t, where x is the horizontal displacement, 3 - We know that for vertical motion, y = ut + ½at² = (Usinθ × t) - ½gt², where y is the vertical height, 4 - Rearrange Equation 2 for t》t = x / Ucosθ, 5 - Substitute Equation 4 into Equation 3》y = (Usinθ × x / Ucosθ) - ½g(x / Ucosθ)², 6 - We know that tanθ = sinθ/cosθ, and 1/cosθ = secθ, 7 - Substitute into Equation 5 》y = xtanθ - (sec²θ)gx²/2u², 8 - We know that sec²θ = 1 + tan²θ, 9 - Substitute into Equation 7 》 y = xtanθ - (gx²(1 + tan²θ)/ 2u²)

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