v m/sec. be the initial speed,

t sec. be the time of flight.

Horizontally, there are no forces on the cannonball, and so it moves with constant speed. Resolving horizontally:

tv cos(60) = 196

t = 196 / v cos(60) …(1)

Vertically, the cannonball has intital velocity v sin(60) upward, and acceleration -g as gravity acts in the downward direction. Its total distance travelled is 0, as it ends up back at ground level.

0 = tv sin(60) – (1/2)gt^2

0 = t(v sin(60) – gt / 2)

t = 0 represents the initial time. You can divide by t, as you are interested in the other (final) time.

0 = v sin(60) – gt/2

gt / 2 = v sin(60)

Substitute for t from (1):

196g / 2v cos(60) = v sin(60)

Multiply by 2v cos(60):

196g = 2v^2 sin(60)cos(60)

v^2 = 196g / 2 sin(60)cos(60)

= 196 * 9.81 / [ 2 * (sqrt(3) / 2) * (1/2) ]

= 196 * 9.81 * 2 / sqrt(3)

v = sqrt(196 * 9.81 * 2 / sqrt(3) )

= 47.12 m/sec.