Using the equation of motion for vertical motion, we have:
\[v_{y}^2 = u_{y}^2 - 2gh\]
where \(v_{y} = 0\) (vertical component of final velocity), \(u_{y} = 14.38 \times \sin(64^\circ)\) (vertical component of initial velocity), \(g = 9.8 \, \text{m/s}^2\)
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