Question

A person jumps from a fourth floor from a height of 15.0m.

The network stretches 1.0m before the person comes to rest.

Determine the average deceleration experienced by the person.

Answer 1

s = - 15 m.     u = 0 m/s     a = - 9.8 m/s/s (gravity)     v = ???

v² = u² + 2as           v

Answer 2

(a) First we need to find the final velocity when the person was falling and the net caught them. Use SUVAT.  

s = - 15 m.     u = 0 m/s     a = - 9.8 m/s/s (gravity)     v = ???

v² = u² + 2as           v

Answer 3

d= 0.5at^2
15=4.9t^2
t= 1.749635531
impact speed = 1.749635531 * 9.8 = 17.1464282 m/s
her average velocity is 8.5732141
d= vt
1= 8.5732141*t
t= 0.116642368
a=v/t
a=17.1464282 / 0.116642368
a= 147 m/s^2
so deceleration is 147 m/s^2