એક ફુગ્ગો 1.25 m / s ^2 ના પ્રવેગથી જમીન

$v =1.25 \times 8 m / s =10\; m / s$

$s =\frac{1}{2} \times 1.25 \times 8 \times 8\; m / s =40 m$

now, $40=-10 t +\frac{1}{2} \times 10 \times t ^2$

$5 t^2-10 t-40=0$

$t^2-2 t-8=0$

hence, $t=4 s$

$s=40 \;m$.

displacement $=40\; m$

Just after being released the stone has an upward velocity, so it will move upwards first.

distance in an upward direction before stopping $d$.

$d =\frac{ v ^2- u ^2}{2 g }=\frac{10^2-0^2}{2 * 10}=5\; m$

and distance $= s +2 d =50 m$

So the distance covered is $50\; m$ and the displacement is $40\; m$.