Question on free fall?

Question

A certain computer hard disk drive is rated to withstand an acceleration of 100g without damage. Assuming the drive decelerates through a distance of 2 mm when its hits the ground, from how high can you drop the drive without ruining it?

Answer 1

The height can be calculated with required knowledge of acceleration due to gravity.

Answer 2

v^2 = 2aS = 2gH; where S = 2E-3 m, g = 9.81 m/sec^2, a = 100g, and H = ? the dropping height assuming the initial velocity u = 0.  v is the impact velocity after falling H distance and it declerates to zero after S = 2E-3 m of deformation.

Solve for H = aS/g = 100gS/g = 100S = 1E2*2E-3 = 2E-1 m = 20 cm.

The physics is this.  The velocity v is generated by the fall from H, while it is stopped by the deformation S.  In other words v^2 = 2gH creates the velocity v and v^2 = 2aS = 2*100gS stops it.

Answer 3

The classic equation that applies here is

s = 1/2 at^2

.002 = 1/2 100*9.81 m/s^2 * t^2

and solving for t implies a deceleration period of 0.002 second.

A deceleration of that magnitude implies a change of speed of

v = a t

v = 100*9.81 * .002 = 1.98 m/s

If the dropped drive is falling at a speed of 1.98 m/s, neglecting air resistance, it must have been falling for

1.98 = 9.81 * t

t=0.2 second

and that means falling over a distance of

s=1/2 a t^2

s = 1/2 9.81 0.2^2 = 0.2 m

20 cm is not very high. Better not drop that drive from a desk then.