A person jumps from a window 34 meters high and is caught in a firefighter's net which stretches 0.6 meter. To the nearest m/s^2, what is the magnitude of the person's acceleration in the net?

Answer 1

v=sqrt(2gH) = sqrt(2•9.8•34) = 25.8 m/s

the decelerated motion
a= [(v(fin) ² -v(init)²]/2•s=(0- 25.8²)/2•0.6=- 555.33≈ - 555 m/s²