Assuming Lin jumps from and returns to the same level ground, the fact that he stays in the air for 0.8 seconds means it takes him 0.4 seconds to reach his maximum height, at which point his vertical velocity will be 0.
Acceleration is a constant [tex]g[/tex] in the downward direction, and when it's constant it's the same as the average acceleration over any given time interval. Over the 0.4 second interval during which he reaches his maximum height, we have
[tex]a_{\rm avg}=-g=\dfrac{0-v_0}{0.4\,\rm s}\implies v_0=3.9\,\dfrac{\rm m}{\rm s}[/tex]
which is also the same speed he will have when he returns to the ground.
To find the maximum height, we have
[tex]0^2-{v_0}^2=2(-g)y_{\rm max}\implies y_{\rm max}=0.78\,\rm m[/tex]
