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For an angle θ\thetaθ measured above the horizontal, how is the horizontal component of a vector of magnitude VVV found?

A

For constant acceleration, velocity is v=u+at\mathbf{v}=\mathbf{u}+\mathbf{a}tv=u+at and position is r=r0+ut+12at2\boxed{\mathbf{r}=\mathbf{r}_0+\mathbf{u}t+\dfrac12\mathbf{a}t^2}r=r0​+ut+21​at2​.

B

The particle is at (x,y)(x,y)(x,y) relative to the origin.

C

$<strong>Vcos⁡θ</strong>\mathbf{<strong>V\cos\theta</strong>}<strong>Vcosθ</strong>$

D

a=(0−g)\mathbf{a}=\begin{pmatrix}0\\-g\end{pmatrix}a=(0−g​)

Card 1 of 22

3.2.10 Extend 1-D techniques to 2-D using vectors (A-level only) Flashcards

  1. A Level
  2. /Maths
  3. /3.2.10 Extend 1-D techniques to 2-D using vectors (A-level only)

22 flashcards on OCR (MEI) A Level Maths 3.2.10 Extend 1-D techniques to 2-D using vectors (A-level only): the key formulae, methods and definitions you need to recall for Component 01, Component 02 and Component 03.

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