Answer
$(a)\space 1.3\space s$
$(b)\space14\space m/s$
Work Step by Step
Please see the attached image first.
(a) First of all let's convert 21.5km/h into m/s by using the conversion factors 1000 m/km, h/3600s as follows.
$21.5\space km/h=(\frac{21.5\space km}{h})(\frac{1000\space m}{km})(\frac{h}{3600\space s})=6\space m/s$
Let's apply the equation $S=ut+\frac{1}{2}at^{2}$ to the package to find the travelled time,
$\downarrow S=ut+\frac{1}{2}at^{2}$
Let's plug known values into this equation.
$8.65\space m= 0+\frac{1}{2}\times 9.8\space m/s^{2}\times t^{2}$
$1.8\space s^{2}=t^{2}$
$t=1.3\space s$
Let's apply equation $S=ut$ to the package to find the horizontal distance.
$\rightarrow S=ut$
Let's plug known values into this equation.
$S=6\space m/s\times1.3\space s$
$S=7.8\space m$
(b) Let's apply $V=u+at$ to the package.
$\downarrow V=u+at$
Let's plug known values into this equation.
$V=0+9.8\space m/s^{2}\times1.3\space s$
$V=12.7\space m/s$
Speed of the package $=\sqrt {(12.7\space m/s)^{2}+(6\space m/s)^{2}}=14\space m/s$
