Solving for Maximum Height of Ball Launched at 60 Degrees

[meagr]

Homework Statement

shot from 60 degrees 128 cm above ground, a metal ball lands 86.5 cm away.
what is max height the ball reaches?

Homework Equations

d=v1 t + 1/2 a tsquared

v2squared = v1squared + 2ad

d= 1/2(v1 + v2)t

a=(v2-v1) / t

The Attempt at a Solution

without having my initial velocity, time, i am having trouble solving for how the max height the ball will reach .
i am trying to solve the vertical displacement being the Vy and the horizontal displacement being the Vx to find out the hypotenuse of the triangle( with angels 60, 30, 90), and i am so lost. i know i have my acceleration due to gravity, and my V2y=0m/s and my angel of elevation is 60 degrees but i am so lost in figuring out max height the ball will reach because i have so many unknowns.

can anyone help?

 

[Delphi51]

Make two headings "horizontal" and "vertical". Write x = vt under the horizontal heading and the accelerated motion formulas under the vertical heading. Fill in all the numbers you have. Note that while you don't know the initial Vx or Vy, you do know their ratio, so enter Vx as an unknown and then Vy as Vx times a tangent. If you have only one unknown in one of the formulas, begin with that! If not, you'll have to use 2 or 3 of them and solve as a system of equations.

Show your equations here if you would like more help!