How to calculate velocity with target 12 m away vertically and 6 m horizontally?

[pill]

I used an online simulation for this lab. The target is 12 m away horizontally and 6 m vertically. I was to find the velocity needed to hit the target.Since it's an online simulation, I played around with it and got 14.3 m/s. The formula I plan on using is:

Vx=range/time
Vy=√(range*10-(range/time)²)
V=√(Vx)+Vy

My problem is, how do I calculate the range and time when the target is hit? Or do I use the final range and time in the equation?

The range and time of the first black X is: 9.19 m and 1 s.
The final range and time is 21.6 m and 2.4 s.

 

[pill]

I was given these equations and was told to solve for V:

t = 12/Vx
Vy = 6/t + (0.5)(9.81)t
Vx = Vcos(50)
Vy = Vsin(50)

And this is what I got:

t = 12/Vx
2.6=12/Vx
Vx=28.8

Vy = 6/t + (0.5)(9.81)t
Vy=6/2.6+(0.5)(9.81)(2.6)
Vy=14.272

Vx = Vcos(50)
28.8/cos(50)=V
V=29.8456

Vy = Vsin(50)
14.272/sin(50)=V
V= -54.395

The velocity from Vy = 6/t + (0.5)(9.81)t is the same from the simulation, but I think I might have messed up in general. I know V from both equations should be the same, but they're obviously not. Any explanation on what I did wrong and how to fix it?

 

[LawrenceC]

What angle with the horizontal is specified? I see the number 50 in your work but don't see where it came from. Without an angle specified, the problem does not have a unique solution.

 

[pill]

What angle with the horizontal is specified? I see the number 50 in your work but don't see where it came from. Without an angle specified, the problem does not have a unique solution.

The angle of the cannon is 50°.

 

[LawrenceC]

Take your equations

t = 12/Vx
Vy = 6/t + (0.5)(9.81)t
Vx = Vcos(50)
Vy = Vsin(50)

and rewrite them with V as one of the unknowns. The other unknown is t. Eliminate t and solve for V. Eliminate the Vx and Vy variables because you know them in terms of V which is what you seek.