I am trying to solve the following:
A man of mass 83 kg jumps down to a concrete patio from a window ledge only 0.48 m above the ground. He neglects to bend his knees on landing, so that his motion is arrested in a distance of about 2.2 cm. What is the average acceleration of the man from the time his feet first touch the patio to the time he is brought fully to rest?
Now, I know that the man has a velocity of $v_1=\sqrt{2gy}$, where $y_1=0.48$ m, when his feet first touch the patio. From the equation of motion for constant acceleration, $a=\frac{v_1^2}{2y_2}$, where $y_2=2.2$ cm. However, I am not sure if I can use this since the equation of motion is derived for constant acceleration. What do I do in the case of average acceleration? Do I need an additional assumption such as constant jerk?