According to the equations of motion for a particle in a straight line under uniform acceleration :
$1)\; v = u + at$
$2)\; s = ut + \frac{1}{2}at^2$
$3)\; v^2 = u^2 + 2as$
$1):$ When ball is thrown directly upwards from the ground
In this case, $u=10\; m/sec,$ $a=g= -10 \;m/s^2$(because I assumed displacement $s$ as positive in upward direction) and $v=0$ at maximum height.
So, from $(1)$, $0= 10 - 10t \Rightarrow t = 1 \;sec$
from $(2)$, $s= 10*1 - \frac{1}{2}*10*1^2 \Rightarrow s = 5 \;m$
$2):$ When ball goes downwards after reaching the maximum height
In this case, $u=0\; m/sec,$ $a=g= +10 \;m/s^2$(because I assumed displacement $s$ as negative in downward direction)
So, from $(3)$, $v^2= 0 + 2*10*5 \Rightarrow v = 10 \;m/sec$
from $(1)$, $10= 0 + 10*t \Rightarrow t = 1 \;sec$
So, from $(1)$ and $(2)$
$1)$ Total time in journey : $1+1 = 2\; sec$
$2)$ Total displacement in journey : $5+5 =\; 10\; m$
$3)$ Velocity of the ball when it hits the ground : $10\;m/sec$