Posted by **Eve** on Monday, September 15, 2008 at 11:39pm.

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?

- Physics -
**Damon**, Tuesday, September 16, 2008 at 5:02am
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

- Physics -
**lakis grete**, Thursday, September 29, 2016 at 12:46am
16.5306

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