We only care about the vertical component of the athlete's position vector, which at time t is
y = (4 m/s) sin(23º) t - 1/2 g t²
where g is the acceleration due to gravity.
The athlete stays airborne for as long as y > 0, which happens for
(4 m/s) sin(23º) t - 1/2 (9.80 m/s²) t² > 0
t (1.56 m/s - (4.9 m/s²) t) > 0
==> 0 < t < (1.56 m/s)/(4.9 m/s²)
==> 0 < t < 0.319 s
so the answer is D.
