Answer
The solution is
$$l=1.26\text{ m}.$$
Work Step by Step
The ball has to cover $h=0.80\text{ m}$ vertical distance moving with the initial speed of $v_{y0}=4.3\sin15^\circ\text{ m/s}$ which means that the final vertical speed is
$$v_{yf}=\sqrt{v_{y0}^2+2gh}$$
which gives for total time
$$t=\frac{v_{yf}-v_{y0}}{g}=\frac{\sqrt{v_{y0}^2+2gh}-v_{y0}}{g}=\\
\frac{\sqrt{(4.3\cdot\sin15^\circ)^2+2\cdot9.81\cdot0.80}-4.3\sin15^\circ}{9.81}\text{ s}=0.30\text{ s}.$$
Since the ball moves horizontally with the constant component of the velocity, the distance covered is
$$l=4.3\text{ m/s }\cdot\cos15^\circ\cdot t=1.26\text{ m}.$$
