Answer:
Step-by-step explanation:
If we assume the car is a point mass in a frictionless world
and v is the initial velocity and the launch is from horizontal ground with no initial vertical velocity
The time to drop 2 m will be
s = ½at²
t = √(2s/a)
t = √(2(2)/9.8)
t = 0.638876...
t = 0.64 s
v = d/t
v = 6 / 0.638876...
v = 9.39148
v = 9.4 m/s