- #1
Matt
A person jumps from a fourth story window 15 m above a safety net. THe jumper stretches the net 1.0 m before coming to rest. What was the deceleration experienced by the jumper?
Equation:
x = x0 + v0t +.5at^2
15 m = 1 m + 8.6 m/s(2s) + .5(a)(1.75s^2)
15m = 1 m + 17.2 m/s^2 + .875s^2(a)
-3.2 m = .875 s^2
-3.66 m/s^2 = Deceleration
Does this look right? I'm not sure about the 1 m as the final position...
Equation:
x = x0 + v0t +.5at^2
15 m = 1 m + 8.6 m/s(2s) + .5(a)(1.75s^2)
15m = 1 m + 17.2 m/s^2 + .875s^2(a)
-3.2 m = .875 s^2
-3.66 m/s^2 = Deceleration
Does this look right? I'm not sure about the 1 m as the final position...