Answer:
[tex]the \: value \: of \: the \: vertical \: component \to : \\ (v_v) = 3m {s}^{ - 1} \\ [/tex]
Step-by-step explanation:
[tex]let \: the \: velocity \: be \to \: v \\let \: the \: horizontal \: component \: be \to \: v_h \\ let \: the \: vertical \: component \: be \to \: v_v \\ applying \: the \: pythagorean \: rule \: of \: vectors \to : \\ {v}^{2} = {v_v}^{2} + {v_h}^{2} \\ {v_v}^{2} = {v}^{2} - {v_h}^{2} \\ {v_v}^{2} = {5}^{2} - {4}^{2} \\ v_v = \sqrt{(25 - 16)} \\ v_v = \sqrt{9} \\ \boxed{ v_v = 3 {ms}^{ - 1} }[/tex]
