Answer
$5.17\space s$
Work Step by Step
Let's apply the equation 3.5b $S=ut+\frac{1}{2}at^{2}$ in the vertical direction to find the flight time of the ball.
$\uparrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation.
$5.5\space m=46sin35^{\circ}m/s\times t+\frac{1}{2}(-9.8\space m/s^{2})t^{2}$
$4.9t^{2}-26.4t+5.5=0$
This is a quadratic equation and using the quadratic formula we can find the value of t.
$t=\frac{-(-26.4\space m/s)\space\pm \sqrt {(-26.4\space m/s)^{2}-4(4.9\space m/s^{2})(5.5\space m)}}{9.8\space m/s^{2}}=\frac{26.4\space m/s\space \pm 24.27\space m/s}{9.8\space m/s^{2}}$
$t=0.214\space s$ or $t=5.173\space s$
The first solution corresponds to the situation where the ball is moving upward and has a displacement of 5.5 m. But this is not the solution we seek. The second solution corresponds to the later time when the ball is moving downward and has a displacement of 5.5 m. So this is our solution for time (t=5.17 s).
