Distance of camera/observer from rocket can be determined by counting how many seconds after explosion one can hear it. Convert to Meters.

Time (t) = to initial explosion = 6.96 (11.96s for later big one)
Speed of sound (s) @ 20c (20 degrees Celsius) = 343m/s
Therefore distance (d) = t x s = 2,387.28 meters (2.38728 km)

Frames with 'UFO' in it:

How many fractions of a second pass between frame 1 and frame 5?
Frame 1 time = 1m 11s 738ms
Frame 5 time = 1m 11s 871ms
Difference = 0.133s

Degrees of arc on screen between both points of where UFO is = approx' 25 degrees.

Now you have two distances, degrees and the time factor - plug into a formula to get the speed. From this image:

You can see that the rocket is nearly perfectly half-way between the two images of the UFO. So we can use simple trigonometry to calculate the distance between the two points (2.4km converted to meters):

Distance traveled = 1.064134 km

Speed = Distance / time

= 1.064317 km / 0.133s
= approx' 8 km/s
= 8 x 60 = km/m = 480 km/m
= 480 x 60 = km/h = approx 28,800 km/h

(if rounded down to 1.06 km and 0.13s speed is 29,353.8461 km/h)

So this was definitely a UFO as bugs, birds, aircraft, etc, don't fly that fast but UFO have been known to do so.