A person jumps from a fourth-story window 15.0 m above a firefighter's safety net. The survivor stretches the net 1.0 m before coming to rest. (a) Wha

Cheyanne Leigh 2020-10-25 Answered

A person jumps from a fourth-story window 15.0 m above a firefighter's safety net. The survivor stretches the net 1.0 m before coming to rest. (a) What was the average deceleration experienced by the survivor when slowed to rest by the net? (b)What would you do to make it "safer" (that is, generate a smaller deceleration): would you stiffen or loosen the net? Explain.

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Theodore Schwartz

Answered 2020-10-26 Author has 99 answers

First, we need to know how fast the person was falling when he hit the net. We know how far he fell, so it is a simple task of determining the his velocity:

v = v_{0} + a t

which means we need to know how long he was falling:

d = v_{0} + \frac{1}{2} a t^{2}

I will assume is initial velocity was zero. We can then solve the above equation for t: (keep in mind d is the distance he fell,and a is gravity, both of these are negative quantities so the negatives cancel)

t = \sqrt{\frac{2 d}{a}} \approx 1.75 s

plugging t into the velocity equation:

v = (- 9.8) (1.75) \approx 17.15 \frac{m}{s}

Now the average deceleration can be found by taking the change in velocity over the change in time. But alas, we do not have tright off hand, so we need something else What we do know is that it took 1m to stop the person. Lets

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Jeffrey Jordon

Answered 2021-10-06 Author has 2027 answers

Lets calculate speed just before hitting net

$v_{0} = \sqrt{2 g h} = \sqrt{2 (9.8 m / s^{2}) (15 m)}$

$v_{0} = 17.15 m / s$

a) Person comes to rest in 1m

$v_{f}^{2} - v_{r}^{2} = 2 a d$

$0^{2} - (17.15)^{2} = 2 (- a) (1 m)$

$a = 147 m / s^{2}$

b) $a = \frac{v_{0}^{2}}{2 d}$

for smaller a ,d have to be larger

so, loosen the net

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