We have equation of motion v = u+at, where v is the final velocity, u is the initial velocity, t is the time taken and a is the acceleration.
Considering vertical motion of megatron
Here u = 0 m/s, a= 9.81 [tex]m/s^2[/tex] and v = 32 m/s
Substituting
32 = 0+9.8*t
t = 3.27 seconds
So it will take 3.27 seconds to reach ground.
Now considering horizontal motion of megatron
We have [tex]s =ut+\frac{1}{2} at^2[/tex], where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.
In this case u = 5 m/s, t = 3.27 seconds, a = 0[tex]m/s^2[/tex]
Substituting
[tex]s = 5*3.27+\frac{1}{2} *0*3.27^2\\ \\ s = 16.35 meter[/tex]
So Megatron falls 16.35 meter away from ground.
