Physics
posted by James on .
An object is thrown vertically and has an upward velocity of 5 m/s when it reaches three fourths of its maximum height above its launch point. What is the initial (launch) speed of the object?

total energy at beginning=totalenergy at point above
1/2 m vi^2=1/2 m 5^2+mg(3/4 h)
this indicates that you have two variables, height, and velocity initially.
but we know in the final 1/4 h, vf=0, and
Vf^2=Vi^2+2gd
0=252*9.8*h/4
or h=50/9.8
now you can solve for vi in the first equation.