Answer:
Acceleration will be equal to [tex]3689.16m/sec^2[/tex]
Explanation:
We have given initial height h = 1 m
Time instant t = 1.2 m-sec = 0.0012 sec
Initial velocity u = 0 m/sec
From third equation of motion we know that [tex]v^2=u^2+2gh[/tex]
So [tex]v^2=0^2+2\times 9.8\times 1[/tex]
v = 4.427 m/sec
We have to find the acceleration
Acceleration is time rate of change of velocity
So acceleration [tex]a=\frac{v}{\Delta t}=\frac{4.427}{0.0012}=3689.16m/sec^2[/tex]
So acceleration will be equal to [tex]3689.16m/sec^2[/tex]