Answer
$v_{y}= 4.80\ m/s.$
Work Step by Step
(c)
(Using the discussion in part (a))
Eq. 4-23 will produce the vertical component at time $t=1.15\mathrm{s}$
$v_{y}=v_{0}\sin\theta_{0}-gt$
$=(25.0m/s) \sin 40.0^{o}-(9.80 m/\mathrm{s}^{2})(1.15s)\\\\=4.80m/s.$
