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The relevant equations are

d=1/2 at2 and

v = at.

Substituting t = v/a into d=1/2 at2 gives

d = v2/2a.

This formula will give the drop height d for any desired speed v.

Plugging in

v = 20mph = 8.9408m/s;

a = 9.8 m/s2 (this is the gravitational acceleration at the earth's surface)

we get

d = 4.078m = 13.38 feet.

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Q: What is the drop height equivalent to a 20 mph collision?
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