Answer
$(a)\space 15.7\space m/s$
$(b)\space 1.6\space s$
Work Step by Step
Please see the attached image first.
(a) To find the final velocity of the package, we can apply the equation $V^{2}= U^{2}+\space2aS$ as follows.
$\downarrow\space V^{2}= U^{2}+\space2aS$
Let's plug known values into this equation.
$V^{2}= 0^{2}+2\times9.8\space m/s^{2}\times 12.5\space m$
$V^{2}= 245\space m^{2}/s^{2}$
$V\space\space = \sqrt {245\space m^{2}/s^{2}}$
$V\space \space= 15.7\space m/s$
So, the package hit the porch at the speed of 15.7 m/s
(b) To find the travel time of the package, we can apply the equation $S= Ut+\frac{1}{2}at^{2}$ to the package as follows.
$\downarrow S= Ut+\frac{1}{2}at^{2}$
Let's plug known values into this equation.
$12.5\space m= (0)t+\frac{1}{2}(9.8\space m/s^{2})\space t^{2}$
$12.5\times2\space m= 9.8\space t^{2}\space m/s^{2}$
$t^{2}=\frac{25}{9.8}\space s^{2}\space =\gt\space t=\sqrt {\frac{25}{9.8}\space s^{2}}=1.6\space s$
Therefore we can write, the package was in the air for 1.6 s
