Answer:
velocity at takeoff = 66.93 [m/s]
Explanation:
For the solution we use the kinematic equation:
[tex]V^{2}=Vi^{2}+2*a*(x-xi) \\ where:\\V=final velocity[m/s]\\Vi=initial velocity[m/s]\\a=acceleration[m/s^{2} ]\\x=rest position [m]\\xi=initial position [m]\\\\[/tex]
Now replacing the data
[tex]V^{2} = 0 + 2*1.6*(1400-0)\\V=66.93[m/s][/tex]
