A particle is projected vertically upwards with an initial velocity of 40 m/s. Find the displacement and distance covered by the particle in 6 s. Take g = 10 $m/s^2$

By first equation the maximum height reached

$v=u+at$

$0=40+(-g)\left(t\right)$

$t=\frac{40}{g}=4$sec

Then, the drop of particle from maximum height

$s=ut+\frac{1}{2}a{t}^2$ $[6sec=4sec+2sec]$

$s=\frac{1}{2}\left(g\right)(2{)}^2=20$m

Maximum height reached by particle

$s=ut+\frac{1}{2}a{t}^2$

$s=40\left(4\right)+\frac{1}{2}\left(10\right)(4{)}^2=40\times 4+5\times 16$

$s=160+80=240$m

Thus,

Displacement $=240-20=220$m

Distance $=240+20=260$m.