Answer:
[tex]v_{i} =28.86\frac{ft}{s}[/tex]
Explanation:
Conceptual analysis
We apply the kinematic formula for an object that moves vertically upwards:
[tex](v_{f} )^{2} =(v_{i} )^{2} -2*g*y[/tex]
Where:
[tex]v_{f}[/tex] : final speed in ft/s
[tex]v_{i}[/tex] : initial speed in ft/s
g: acceleration due to gravity in ft/s²
y: vertical position at any time in ft
Known data
For [tex]v_{f} = 25\frac{ft}{s}[/tex] ,[tex]y=\frac{1}{4} h[/tex]; where h is the maximum height
for y=h, [tex]v_{f} =0[/tex]
Problem development
We replace [tex]v_{f} = 25\frac{ft}{s}[/tex] , [tex]y=\frac{1}{4} h (ft)[/tex] in the formula (1),
[[tex]25^{2} =(v_{i} )^{2} -2*g*\frac{h}{4}[/tex] Equation (1)
in maximum height(h): [tex]v_{f} =0[/tex], Then we replace in formula (1):
[tex]0=(v_{i} )^{2} - 2*g*h[/tex]
[tex]2*g*h=(v_{i} )^{2}[/tex]
[tex]h=\frac{(v_{i})^{2} }{2g}[/tex] Equation(2)
We replace (h) of Equation(2) in the Equation (1) :
[tex]25^{2} =(v_{i} )^{2} -2g\frac{\frac{(v_{i})^{2} }{2g} }{4}[/tex]
[tex]25^{2} =(v_{i} )^{2} -\frac{(v_{i})^{2} }{4}[/tex]
[tex]25^{2} =\frac{3}{4} (v_{i} )^{2}[/tex]
[tex]v_{i} =\sqrt{\frac{25^{2}*4 }{3} }[/tex]
[tex]v_{i} =28.86\frac{ft}{s}[/tex]
