How can I solve this projectile motion problem?

if it traveled 47 yards in 5.3 seconds, you know its horizontal velocity (which is constant during flight when you neglect friction ) is 47yds/5.3s=8.87yds/s = 8.11m/s

if the hang time was 5.3s, it took half of that, or 2.65 secs to reach max altitude; since we know vertical velocity =0 at max alt, and we know gravity reduces vertical speed by 9.8m/s for each sec of flight, we know that the object had an initial vertical speed of 2.65s*9.8m/s/s=26m/s

now, if the initial velocity had a vertical component of 26m/s and the horizontal had a component of 8.1m/s, we find the launch angle from tan(theta)=vy/vx=26/8.1 => theta = 72.7 deg

the max alt can be determined by realizing the ball fell from a height in a time of 2.65s, so we have

h=1/2 gt^2 = 1/2 * 9.8m/s/s*2.65^2 = 34.5m

we can also find this from

vfy^2=v0y^2+2ad

vfy=final vert velocity =0
v0y=initial vert velocity = 26m/s
a=g=-9.8m/s/s
d=distance traveled

0=26^2-2(9.8)d => d= 26^2/(2*9.8)=34.5m