Question

A ball dropped from some height covers half of its total height during the last second of its free fall. Find
(a) time of flight
(b) height of its fall
(c) speed wit which it strikes the ground.

Answer 1

Time of free fall=n seconds

Height of free fall =H

$H_n=\frac{1}{2}g{t}^2=\frac{1}{2}g{n}^2$

Similarly,$H(n-1)=\frac{1}{2}g(n-1{)}^2$

Distance traveled in last second $=\frac{1}{2}g[n^2-(n-1/2{)}^2]=\frac{1}{2}g(2n-1)$

As per the question $h=\frac{1}{2}{h}_n$

$\frac{1}{2}g(2n-1)=\frac{1}{4}g{n}^2$

$n^2-4n+2=0$

$n=(2+\sqrt{2}),(2-\sqrt{2})$

Since, $(2-\sqrt{2})$ is less than 1 it cannot be taken

(i)$n=(2+\sqrt{2})$

(ii)Height of the free fall $=\frac{1}{2}g(2+\sqrt{2}{)}^2=58.2m$

(iii)$v_f=u+gt$

$=10(2+\sqrt{2})=34.14m/s$