Physics Insight

Question pattern · Knowledge card

Projectile launched above horizontal

Resolve the launch velocity first, then follow horizontal and vertical components with a consistent sign convention and the physically valid flight-time root.

Primary skill
Calculate & apply
Original practice questions
2

Connect the ideas

Key relationships and graph reading

  • uₓ = u cos θ and uᵧ = u sin θ
  • At maximum height vᵧ = 0
  • For equal launch and landing height without resistance, flight time = 2uᵧ / g

Check your method

Three common mistakes

  1. Using the full launch speed on both axes.
  2. Assuming equal ascent and descent times when landing height differs.
  3. Keeping a negative mathematical time root.

Original micro-example

Try the idea in miniature

A projectile is launched at 10 m s⁻¹, 30° above horizontal, and lands at its launch height. Use g = 9.8 m s⁻².

Show the short answer

uᵧ = 5.0 m s⁻¹, so the flight time is 2(5.0)/9.8 ≈ 1.02 s. The horizontal component is 8.66 m s⁻¹.

Original practice questions

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