Max height-2 balls-1 straight up-the other horizontal at an angle? yikes

[sddatt]

Homework Statement

A ball is thrown straight upward and returns
to the thrower’s hand after 4 s in the air. A
second ball is thrown at an angle of 67◦ with
the horizontal.
The acceleration of gravity is 9.8 m/s2 .
At what speed must the second ball be
thrown so that it reaches the same height as
the one thrown vertically? Answer in units of
m/s.

The Attempt at a Solution

*****I've been working on this problem for a few days. Here's what I have so far:

This is a 2-d question. First I need to find the maximum height of the first ball (thrown straight up) then I can plug that into the second ball's equations and find the Vo(x) of the second ball.
**Will the ball that's being thrown straight up have it's maximum height at 2s (half way between the total time?)
So I would have Y=0 + 0(1)+ (+)4.9(1^2).
Then Height would be +4.9? I think that gravity COULD be + because the ball is going up...but I don't know if that even makes sense.

Then I would put the Y into my second ball's equations:
1.) Y EQN: 4.9=0+Vo(y)sin(67)(t)+4.9(t^2)
2.) X EQN: x=0 + Vo(x)cos(67)(t)+ 0
3.) Vy EQN: 0=Vo(y) + -9.8(t)
4.) Vx EQN: Vx=Vo(x) + 0

**Substitute and solve for t: t=.___s
**Plug t into X EQN: x=Vo(x)cos(37)(.__s)
Now I am missing x and the magnitude.

I really don't know what to do with this problem. Any help is greatly appreciated! Thank you! :)

 

[Kurdt]

You need to find the initial vertical velocity the first ball has. That will be the same vertical velocity your second ball has.