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 January 15th, 2019, 06:53 AM #1 Senior Member   Joined: May 2014 From: Allentown PA USA Posts: 110 Thanks: 6 Math Focus: dynamical systen theory How is this answer possible? Dear My Math Forum: A ball of clay falls to the ground from a height of 15 meters. It is in contact with the ground for 0.020 seconds before coming to rest. What is the average acceleration of the clay during the time it is in contact with the ground? The correct answer to this problem is 858 meters/second^2. Could someone please provide me with a step-by-step demonstration as to how this answer was arrived at? Thank you.
 January 15th, 2019, 07:17 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,885 Thanks: 1504 $\bar{a} = \dfrac{\Delta v}{\Delta t}$ $\Delta v$ = (rest velocity) - (final velocity at the start of impact) note ... $v_f^2 = v_0^2 - 2g \Delta y \implies v_f = -\sqrt{0 - 2g(-15)} = -\sqrt{30g}$ $\bar{a} = \dfrac{0 - (-\sqrt{30g})}{0.02}$ using $g = 9.81 \, m/s^2$ yields the average acceleration given.

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