A particle is projected vertically upwards with velocity $40{ms}^{-1}$. Find the displacement and distance travelled by the particle in 6 s.
[take $g=10m/s^2$]

The correct option is B $60m,100m$
Initial upward velocity $u=40$ $m{s}^{-1}$

Let after time $t$, the particle reaches the maximum height.

So final velocity $v=0$

Acceleration $a=-g$

Using $v=u-gt$

$\therefore$ $0=40-10t$ $⟹t=4$ $s$

Thus after $4$ $s$, the particle starts the downward motion for another $2$ $s$

• Distance traveled in upward motion:

Here we take $a=-g$ as it is downward (negative) and distance is upwards (positive)

$S_1=ut-\frac{1}{2}g{t}^2$ where $t=4$ $s$

$\therefore$ $S_1=40\times 4-\frac{1}{2}10\times 4^2=80$ $m$

• Distance traveled in downward motion

Here we take $a=g$ as it is downward (positive) and distance is also (positive)

$S_2=u^{\prime }t+\frac{1}{2}g{t}^2$ where $t=2$ $s$ and $u^{\prime }=0$

$\therefore$ $S_2=0+\frac{1}{2}10\times 2^2=20$ $m$

• Total distance covered in $6s$: $S_T=80+20=100$ $m$

• Displacement in $6$ $s$:

$S=ut-\frac{1}{2}g{t}^2$ where $t=6$ $s$

$S=40\times 6-\frac{1}{2}10\times 6^2=60$ $m$

Option A is the answer.