V^2 =u^2 + 2asV =0 u = ? a = 9.8 s = lets say 1.5m
V^2 = 2 x 9.8 x 1.5
V = 5.4 ms
V= u + at
5.4 = 9.8
So it'll fall to the ground in 0.55 seconds.
Which means it'll have to travel at 581,000 feet per second to travel 320,000 feet.
If he can't dodge Freeza's beam from 20 feet away, how fast does the beam have to travel?
Thanks. We could make a calculating super genius team, you'll be the calculator/super genius, and I'll be the guy who receives the credit and money.😄 JK
I'm going to assume the beam travels 20 times as fast as Goku, if Goku can travel twice as far as the rubble in half the time, then he can travel 4 times faster than the rubble was in the first 5 seconds, thats 2,324,000 feet per second, so the beam travels at 46,480,000 feet per second, if you look at the four beams and Goku's after images in Vo. 11, PG 87, there are four beams coming at Goku, he ducks under the first one, and stands back up, and then moves to the right of the second one, which was 2 feet to the right of the first one, and a little bit higher, (NOTE, Goku's dodging them like this to make it known that Freeza no longer has control over the fight.) and then he moves to the left of the third one, which was about 5 feet to the left of the second, and then he moves to the left of the fourth one, which was in between the first and the third.