d = 1.50 m ( max height )
vf² = vi² + 2 a d
where: vf is the final velocity, vi is the initial velocity and a = - g = - 9.81 m/s²
vf = 0
0² = vi² - 2 · 9.81 · 1.50
vi² = 29.43
vi = √29.43
vi = 5.42 m/s
Answer: His velocity as he leaves the water must be 5.42 m/s.