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3.2.8 Motion under gravity using vectors (A-level only)

Card 1 of 21

With upward positive, what is the acceleration vector under gravity?

A

a=(0−g)\mathbf{a}=\begin{pmatrix}0\\-g\end{pmatrix}a=(0−g​), where g=9.8 m s−2g=9.8\text{ m s}^{-2}g=9.8 m s−2.

B

Substitute the same ttt into the horizontal position equation.

C

The particle is moving downwards.

D

When launch and landing are at different vertical levels.

Card 1 of 21

3.2.8 Motion under gravity using vectors (A-level only) Flashcards

  1. A Level
  2. /Maths
  3. /3.2.8 Motion under gravity using vectors (A-level only)

21 flashcards on OCR A Level Maths 3.2.8 Motion under gravity using vectors (A-level only): the key formulae, methods and definitions you need to recall for Paper 1, Paper 2 and Paper 3.

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