Gravity (g=9.81m/s/s approx) causes the ball to follow a parabolic trajectory (assuming negligible air resistance) and s=u(y)t-½gt^2, where t=1.8s is the time in the air, u(y) is the vertical component of the ball's velocity and s is the vertical distance at the final destination of the ball, so s=h (where h is the height of the ball when it's caught).
The horizontal distance=u(x)t where u(x) is the horizontal component of the ball's velocity. So 43.3=1.8u(x) and u(x)=43.3/1.8=24.06m/s approx.
The balls's velocity is √((u(y))^2+(u(x))^2)=26. We can find u(y)=√676-24.06^2=9.87m/s approx.
We can now find out how long it would have taken for the ball to fall to the ground if it hadn't been caught. The final height of the ball would be zero, so 0=u(y)t-½gt^2, and t=0 or 2u(y)/g=2*9.87/9.81=2.011. The ball reaches its maximum height at half this time=1.01s approx. The actual height is 9.87*1.01-4.905*1.02=4.95m approx.
The height of the ball when caught was about 1.87m=9.87*1.8-½g*3.24.
The question is slightly ambiguous: it could mean the maximum height (4.95m) of the ball before it's caught or the height it was caught (1.87m).