At the ball's highest point, it has no vertical velocity, so the 6 m/s is purely horizontal. A projectile's horizontal velocity does not change, which means the ball was initially thrown with speed <em>v</em> such that
<em>v</em> cos(53°) = 6 m/s ==> <em>v</em> = (6 m/s) sec(53°) ≈ 9.97 m/s
The player shoots the ball from a height of 2.0 m, so that the ball's horizontal and vertical positions, respectively <em>x</em> and <em>y</em>, at time <em>t</em> are
<em>x</em> = (9.97 m/s) cos(53°) <em>t</em> = (6 m/s) <em>t</em>
<em>y</em> = 2.0 m + (9.97 m/s) sin(53°) <em>t</em> - 1/2 <em>gt</em> ²
Find the times <em>t</em> for which the ball reaches a height of 3.00 m:
3.00 m = 2.0 m + (9.97 m/s) sin(53°) <em>t</em> - 1/2 <em>gt</em> ²
==> <em>t</em> ≈ 0.137 s or <em>t</em> ≈ 1.49 s
The second time is the one we care about, because it's the one for which the ball would be falling into the basket.
Now find the distance <em>x</em> traveled by the ball after this time:
<em>x</em> = (6 m/s) (1.49 s) ≈ 8.93 m