WAEC 2013 · Paper 2 · Q8

Kinematics & dynamics

A stone is thrown vertically downwards from the top of a tower of height $45\text{ m}$ with a speed of $20\text{ m s}^{-1}$ . Find the: $[\text{Take } g = 10\text{ m s}^{-2}]$

(a)
time it takes to reach the ground;
(b)
speed with which it hits the ground.

Worked solution (try it first)

Take down as positive: $u = 20$ , $a = 10$ and $s = 45$ .

(a)

$s = ut + \frac12at^2$ : $45 = 20t + 5t^2$ , so $t^2 + 4t - 9 = 0$ .
$t = \dfrac{-4 + \sqrt{16 + 36}}{2}$
$= -2 + \sqrt{13}$
$\approx 1.61\text{ s}$ (the other root is negative).

(b)

$v^2 = u^2 + 2as = 400 + 900 = 1300$ , so $v = \sqrt{1300}$
$\approx 36.06\text{ m s}^{-1}$ .

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