WAEC 2017 · Paper 2 · Q14

(a)
A particle is projected vertically upwards from a point $O$ with a velocity of $75\text{ m s}^{-1}$ . Find the: (i) velocity of the particle at the end of 5 seconds; (ii) height attained when the velocity is $15\text{ m s}^{-1}$ ; (iii) times when the particle is $270\text{ m}$ above $O$ . $[\text{Take } g = 10\text{ m s}^{-2}]$
Separate values with commas, e.g. 3, −2
(b)
A force of $24\text{ N}$ acts on a mass $q\text{ kg}$ and increases its speed from $6\text{ m s}^{-1}$ to $42\text{ m s}^{-1}$ in 6 seconds. Find the value of $q$ .

Worked solution (try it first)

$v^2 = u^2 - 2gs$ : $15^2 = 75^2 - 20s$ , so $s = \dfrac{5625 - 225}{20} = 270\text{ m}$ .

$270 = 75t - 5t^2$ , so $t^2 - 15t + 54 = 0$ .
$(t - 6)(t - 9) = 0$ : at $t = 6\text{ s}$ (going up) and $t = 9\text{ s}$ (coming down).