Answer
(a) $0.616s$
(b) $0.79m$
Work Step by Step
(a) We know that
$v_{\circ y}=v_{\circ}sin\theta$
$\implies v_{\circ y}=(2.62)sin60.5^{\circ}=2.28\frac{m}{s}$
and $v_y^2=v_{\circ y}^2-2g\Delta y$
$\implies v_y=\sqrt{v_{\circ y}^2-2g\Delta y}$
We plug in the known values to obtain:
$v_y=\pm \sqrt{(2.28)^2-2(9.81)(-0.455)}$
$v_y=-3.76\frac{m}{s}$
Now we can find the required time:
$t=\frac{v_y-v_{\circ y}}{-g}$
$\implies t=\frac{-3.76-2.28}{-9.81}$
$t=0.616s$
(b) The required horizontal distance can be determined as:
$x=(v_{\circ}cos\theta)t$
We plug in the known values to obtain:
$x=(2.62)cos 60.5^{\circ}(0.616)$
$x=0.79m$
