Question

During the last second of its free fall, a body covers half of the total distance travelled. Calculate (i) the approximate height from which the body falls (ii) the duration of the fall.

Answer 1

$\frac{h}{2}=\frac{1}{2}g(t-1{)}^2$ ...(1)
$h=\frac{1}{2}g{t}^2$ ...(2)
$t^2+1-2t=\frac{t^2}{t}⟹t^2-4t+2=0$
$g(t-1{)}^2=\frac{g{t}^2}{2}$
$t=\frac{4\pm \sqrt{16-8}}{2}=\frac{4\pm 2\sqrt{2}}{2}$
$t=2\pm \sqrt{2}$
Now $t-1>0$ if $t=2-\sqrt{2}$
$t-1=2\sqrt{2}-1=1-\sqrt{2}<0$
(ii) So $t=2+\sqrt{2}$
(i) $h=\frac{1}{2}g{t}^2=17m$