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?

Answer 1

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

Answer 2

16.5306