Finding Maximum Height and Initial Velocity of a Thrown Ball

[Bkrdc04]

Homework Statement

Standing on the ground, you throw a baseball straight upward, releasing the ball at a height of 2.00 meters. The ball travels straight up and then falls(ignoring air resistance). If the ball was in the air for a total of 3.92s, find the maximum height the ball reaches and the speed at which it was thrown.

Homework Equations

Vf=Vi+at
(Vf^2)-(Vi^2)=2ax
x=(Vi*t)+(1/2)(a)(t^2)

The Attempt at a Solution

I divided my diagram into 3 velocities:
velocity initial (Vo):when the ball was released at 2.00 meters
velocity 2 (V2): velocity at the top of "bell" which is zero
and the velocity once it hit the ground (Vf)
Vi=? V2=0 m/s Vf=? i assume Vf is not important

I am treating time as a net total (3.92) and therefore tnet=3.92=t1+t2
t1: being the time it take for the ball to reach v2
t2: being the time it takes for the ball to go from v2(top of the bell with 0m/s) to the ground

*I treated x the same. Xnet=?=x1+x2
xi:the distance the projectile traveled until velocity became zero. I am assuming x1=2+?
x2: the distance the projectile traveled before reaching x=0 meters

I played around so much with the equations (substituting equations into each other) that I kept getting 2*G as the acceleration. I know that's not right because the ball was thrown upwards and therefore the velocity must be positive.

The answer ended up being:

the ball reached 19.8m high
and 18.7 m/s as the initial velocity.
I've never had a problem like this that hasn't included an initial velocity.

 

[PhanthomJay]

You've got the right equations; try your third one to solve for Vi (what is x and what is a and what is t?). Then use your 2nd equation to solve for h.