Chapter 5 - Newton's Laws of Motion - Problems and Conceptual Exercises - Page 146: 76

$F=\frac{-mv^2}{2\Delta x}$

We know that $v_f^2-v_i^2=2a\Delta x$ As $v_f=0$, $v_i=v$ $\implies (0)^2-v^2=2a\Delta x$ $\implies a=\frac{-v^2}{2\Delta x}$ Now $F=ma$ $\implies F=m(\frac{-v^2}{2\Delta x})$ $F=\frac{-mv^2}{2\Delta x}$ The negative sign shows that the force is opposite to the motion of the ball.