Stuck on projectile question please help if you have any idea?

Question

The question is that a mass is shot from a cannon and the end of the cannon's barrel is at a height of 6.9m. The initial velocity vi of the projectile has a horizontal component of 4.6m/s. The projectile rises to a maximum height of delta y above the end of the cannon's barrel and strikes the ground a horizontal distance delta x past the end of the cannon's barrel and the acceleration of gravity is 9.8m/s^2. I have to find the time it takes for the projectile to reach its maximum height in seconds. I used vy^2=voy^2-2g(y-yo) and I get 7.6 but it's not the right answer if someone knows what I did wrong that would be so helpful.

Answer 1

There's not enough info. You need either:
elevation angle
v0y
delta y
y (max)
time of flight
hor. distance of flight
You mentioned v0y and y in your question. Do you have either of those values?
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OK, you emailed me that the angle is 57 deg. Now we can figure out v0y and v0 and plug the data into a comprehensive formula provided by the ref.
v0 = v0x/cos(57deg) = 8.44596091037265 m/s. (I assume this is a toy cannon.)
The formula assumes a 0 altitude launch point and allows any altitude target height yt. In this case we specify yt = -6.9 m and must add 6.9 m to any y data resulting.
v0x, voy = cos,sin(theta)*v0 = 4.60000000000001 7.08337883354708
Flight time (time when y=yt=-6.9) =
v0[y]/g + sqrt((v0[y]/g)^2-2yt/g) = 2.11225195605011
Range at y=yt=-6.9 = v0[x]*time when y=yt = 9.7163589978305
ymax = 2.55991100507871 (+6.9 = 9.45991100507871)
Time when y=ymax = sqrt((2ymax/g) = .722793758525212
Range at y=ymax = v0[x]*time when y=ymax = 3.32485128921598