Question

For comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a height of 1.0 m and stopped in a 1.2-ms-long encounter with a hard floor?

Answer 1

Answer:

Acceleration will be equal to $3689.16m/sec^2$

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 $v^2=u^2+2gh$

So $v^2=0^2+2\times 9.8\times 1$

v = 4.427 m/sec

We have to find the acceleration

Acceleration is time rate of change of velocity

So acceleration $a=\frac{v}{\Delta t}=\frac{4.427}{0.0012}=3689.16m/sec^2$

So acceleration will be equal to $3689.16m/sec^2$