Physics
posted by Eve on .
An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one fourth of its maximum height above its launch point. What is the initial (launch) speed of the object?

v = Vo  9.8 t
height versus time is a parabola
h = Vo t  4.9 t^2
Vertex is at the top where v = 0 call that height capital H
0 = Vo  9.8 ttop
ttop = Vo/9.8
H = Vo ttop  4.9 ttop^2
H = Vo^2/9.8  .5 Vo^2/9.8
H = .5 (Vo^2/9.8)
H/4 = .125 (Vo^2/9.8)
at some time t it is at H/4 and v =18
v = 18 = Vo  9.8 t
H/4 = Vo t  4.9 t^2 = .125 Vo^2/9.8
Vo = 18 + 9.8 t
that gives two equations in the two unknowns, t and Vo
solve for Vo 
16.5306