Question

A body projected vertically up has displacement 20 m in the first 'n' seconds while it was moving up. Find its magnitude of displacement in the last 'n' seconds while falling down.

Answer 1

$s=ut-\frac{1}{2}g{t}^2$

$20=u.n-\frac{1}{2}g{n}^2$..(1)

Time of flight $=\frac{2u}{g}$

$n+t_0+n=\frac{2u}{g}$

$t_0=\frac{2u}{g}-2n$

$s=u(n+t_0)-\frac{1}{2}g(n+t_0^2)$

$=u(n+\frac{2u}{g}-2n)-\frac{g}{2}(n+\frac{2u}{g}-2n{)}^2$

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

$=\frac{2u^2}{g}-un-\frac{g}{2}(\frac{4u^2}{g^2}+n^2-\frac{4un}{g})$

$=\frac{2u^2}{g}-un-\frac{2u^2}{g}-\frac{n^{2}g}{2}+2un$

$s=un-\frac{n^{2}g}{2}$ ..(2)

comparing (1) and (2)

$s=20m$

Magnitude and displacement in the last 'n' seconds =20 m