- #1

Dorothy Weglend

- 247

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So, obviously, a=3 m/s^2, since we want F=(80 kg)(3 m/s^2). I use v^2 = 2g(1.5) to get the velocity at the point that the knees start to bend, so v^2 = 29.4 (m^2/s^2).

Since I want a=3 m/s^2, I reason that all I need is to use 0 = v^2 - 2ah, with a=3 and v^2=29.4, to find the stopping distance at that acceleration, which should be h.

This gives me 4.9 m, which seems excessive, and French gives h = 0.75 m, and I can't see how he ended up with that. It seems to me this would correspond to a final velocity of 2.12 m/s, which is way too low for a 1.5 m drop.

Well, thank you for any suggestions or pointers.

Dot