A particle has a speed of 6 ms⁻¹ in a direction relative to unit vectors i and j as shown in the diagram below - AQA - A-Level Maths Pure - Question 12 - 2021 - Paper 2

To find the vertical component of the particle's velocity, we can use the sine function. Given the angle of 30 degrees and the hypotenuse (the speed) of 6 ms⁻¹, the expression for $v_2$ (the vertical component) is:

$v_2 = 6 imes ext{sin}(30^ ext{o})$

Using the fact that $ext{sin}(30^ ext{o}) = 0.5$, we can simplify this to:

$v_2 = 6 imes 0.5 = 3 ext{ ms}^{-1}$

Thus, the correct expression is:

$v_2 = 6 ext{ sin } 30^ ext{o}$