Ordinary scatter

The initial velocity of ball is $v_i = 10\hat{i} - 10\hat{j}$ . Let $v_y$ and $v_x$ be the final components of velocity of ball along y-axis and x-axis.

The impulse from the ground acts in vertical direction. Hence the vertical component of velocity will change.The component of velocity of ball along x-axis after impact will remain unchanged since no impulse acts on the ball along x-axis. Hence $v_x$ =10 .

Along vertical,

$\begin{aligned} e &= \frac{\text{velocity of separation}}{\text{velocity of approach}} = \frac{v_y}{10} \\ \Rightarrow v_y &= 5 \\\\ v_f &= 10\hat{i} + 5\hat{j} \\ | \Delta p | &=m \times (|v_f - v_i|) \\ &=10 \times |15j| \\\\ &=150. \end{aligned}$

By impulse momentum theorem , the change in momentum is given by the integral of force with respect to time $\begin{aligned} \int F dt &= \langle F \rangle \Delta t \\ \langle F \rangle &= \frac{| \Delta p| }{\Delta t} = \frac{ 150}{0.1} \\ \langle F \rangle &=\boxed{\SI[per-mode=symbol]{1500}{\newton}} \end{aligned}$