WAEC 2011 · Paper 2 · Q17

(a)
A particle is projected vertically upwards with a speed of $25\text{ m s}^{-1}$ from a point on the ground. Find the: (i) position of the particle after 4 seconds; (ii) maximum height reached; (iii) time taken to reach the maximum height; (iv) times when the particle is $30\text{ m}$ above the ground. $[\text{Take } g = 10\text{ m s}^{-2}]$
Separate values with commas, e.g. 3, −2
(b)
Calculate the force which acts on a body of mass $3\text{ kg}$ moving at $2.5\text{ m s}^{-1}$ for $0.5$ seconds, if the final velocity is $4.5\text{ m s}^{-1}$ .

Worked solution (try it first)

$s = ut - \frac12gt^2 = 25(4) - 5(16) = 20$ , so it is $20\text{ m}$ above the ground.

At the top $v = 0$ : $0 = 25^2 - 20s$ , so $s = \dfrac{625}{20} = 31.25\text{ m}$ .

$30 = 25t - 5t^2$ , so $t^2 - 5t + 6 = 0$ .
$(t - 2)(t - 3) = 0$ : at $t = 2\text{ s}$ (going up) and $t = 3\text{ s}$ (coming down).