Answer
(a) $0.199s$
(b) $0.452m$
Work Step by Step
(a) We know that
$v_{\circ y}=v_{\circ}sin\theta$
$\implies v_{\circ y}=(2.62)sin(-30.0^{\circ})=-1.31\frac{m}{s}$
$v_y^2=v_{\circ y}^2-2g\Delta y$
$\implies v_y=\pm \sqrt{(-1.31)^2-2(9.81)(-0.455)}=-3.26\frac{m}{s}$
Now we can find the required time as
$t=\frac{v_y-v_{\circ}y}{-g}$
We plug in the known values to obtain:
$t=\frac{-3.26-(-1.31)}{-9.81}$
$t=0.199s$
(b) The required horizontal distance can be determined as:
$x=(v_{\circ}cos\theta)t$
We plug in the known values to obtain:
$x=(2.62)cos(-30.0^{\circ})(0.199)$
$x=0.452m$
