A Nice Blend of Newton's laws and kinematics

The force exerted by the object on the ground $= mg = 5 \times 10 = 50N$ .The force that the block is exerted from downwards $= 100 N$ .Hence we have :

$F_{net} = 100 - 50 = 50N \ (upwards)$

Since $F=m.a_1$ we have :

$a_1=\dfrac{F_{net}}{m} = \dfrac{50}{5} \\ \Rightarrow a_1 = 10m.s^{-2}$

By Kinematic equation we have:

$s_1=u_1t_1+\dfrac{1}{2}a_1t_1^2\\ \Rightarrow s_1=(0)(10)+\dfrac{1}{2}(10)(10)^2 \\ \Rightarrow s_1=500m$

Now the upward acceleration of the block stops but it still continues to move due to the Law of Inertia.Hence we must calculate the velocity of the block at the instant where the acceleration stops.Denote it by $v$ .Note that we will use this $v$ as $u_2$ of second part.

So by Kinematic Equation we have:

$v=u_1+a_1t_1 \\ \Rightarrow v=0 + (10)(10) \\ \Rightarrow v=100m.s^{-1}$

Again by Kinematic equation we have:

$v_2^2=u_2^2-2.(a_2).(s_2) \\ \Rightarrow 0 = 100^2 - 2(10)s_2 \\ \Rightarrow s_2=\dfrac{10000}{20} \\ \Rightarrow s_2=500m$

Hence , the total height reached by the block is :

$s_1+s_2=500m+500m=\Large\boxed{1000m}$