Hello
This is a problem of accelerated motion, where the acceleration involved is the gravitational acceleration: [tex]g=-9.81~m/s^2[/tex], and where the negative sign means it points downwards, against the direction of the motion.
Therefore, we can use the following formula to solve the problem:
[tex]v_f^2 = v_i^2 + 2gS[/tex]
where [tex]v_i=4~m/s[/tex] is the initial vertical velocity of the athlete, [tex]v_f=0[/tex] is the vertical velocity of the athlete at the maximum height (and [tex]v_f=0~m/s[/tex] at maximum height of an accelerated motion) and S is the distance covered between the initial and final moment (i.e., it is the maximum height). Re-arranging the equation, we get
[tex]S= \frac{v_f^2-v_i^2}{2g}=0.82~m [/tex]
