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October 30, 2014

October 30, 2014

Posted by **johnny** on Thursday, December 11, 2008 at 8:04pm.

(a) What is his average acceleration from when his feet first touch the patio to when he stops?

m/s2

(b) What is the magnitude of the average stopping force?

kN

- physics -
**bobpursley**, Thursday, December 11, 2008 at 8:11pmI will be happy to critique your thinking.

- physics -
**Christiaan**, Thursday, December 11, 2008 at 8:14pma) First you calculate his speed of impact (the speed when he hits the ground).

For this you can use the formula:

v² = v0² + 2.a.x

where:

-v is your final speed

-v0 is your initial speed (0 m/s in this case)

-a is your acceleration (the gravitational acceleration in this case, being 9.81 m/s²)

-x, the distance travelled (being 0.6 m in this case)

so v² = 2*9.81 m/s²*0.6 m = 11.77 m²/s²

Now, we can use the same formula to calculate his average acceleration when he is slowing down.

v² = v0² + 2*a*x

- v being 0 m/s in this case

- v0² being 11.77 m²/s² (since he starts slowing down at this speed)

- x, being 0.02m

=> a = (v² - v0²)/(2*x) = (-11.77 m²/s²)/(2*0.02m) = - 294,25 m/s²

- physics -
**Christiaan**, Thursday, December 11, 2008 at 8:17pmb) Since F = m*a

The average stopping force can be calculated as:

F = 72 kg* -294.25 m/s² = -21.186 kN

The negative sign for both the acceleration and stopping force means they are poiting upward, since we took the downward direction to be positive.

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