Projectile motion of mini cannon ball

[Thunderer]

I am not too sure whether I am doing this correctly.

The problem is basically a mini-cannon aimed at a 60 degree angle, hits a target on the ground 0.8m away. I am suppose to find the initial velocity.

This is what I did:
Cut the distance in half, assuming the mini cannon-ball will reach its maximum height at that point (and v = 0). Which is 0.4 m. I then use trig, and do the following: tan60 = y/0.4; y = 0.693m

vf^2 - vi^2 = 2a(xf-xi)
vf^2 - 0 = 2(-9.8m/sT=^2)(0.693m)
vf = 3.68 m/s from top of height to ground
so vi = 3.68 m/s
So this is the velocity of the y component?

Then:
tan 60 = (3.68 m/s) / x
xtan(60) = 3.68 m/s
x = 3.68/tan60 = 2.12 m/s
v(total)^2 = vy^2 + vx^2 = (3.68 m/s)^2 + (2.12 m/s)^2
v(total) = 4.25 m/s?

Is that correct? Did I do anything wrong?

[I also did this, but I don't think I need it.
vf = vi + at
0 = 3.68 m/s + (-9.8 m/s^2)(t)
t = 0.376 s ]
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[J Hann]

The formula for the distance of a projectile with no air resistance is

s = v^2 * sin(2 * theta) /g so v = (s * g / sin(2 * theta)^1/2
s = (.8 * 9.8 / .867)^1/2 = 3 m/s

The time for the projectile to reach max height is v * sin(theta) /g
since vf = vi - gt and vf = 0 at the top of the trajectory
Multiply this by 2 to get total time of flight and t = 2 * v * sin(theta) / g
Also, the total distance traveled is s = v * cos(theta) * t
Substitute the value for t and s = v^2 * 2 * sin(theta) * cos(theta) /g
Since 2 * sin(theta) * cos(theta) = sin(2 * theta) you get the formula
for the distance s =v^2 * sin(theta) / g as stated above

 

[Doc Al]

This is what I did:
Cut the distance in half, assuming the mini cannon-ball will reach its maximum height at that point (and v = 0). Which is 0.4 m. I then use trig, and do the following: tan60 = y/0.4; y = 0.693m

It is certainly true that the cannonball will reach its maximum height at x = 0.4 m. But it is not true that tan60 = y/0.4. The initial angle of the cannonball's trajectory is 60 degrees, but it moves in a parabolic arc not in a straight line.