Answer to Question #202052 in Optics for Tshephang

Question

 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, Fig. 3. 

(a) What was the average deceleration experienced by the survivor when she was slowed 

to rest by the net? (5)

Answer 1

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

Given quantities:

s = - 15 m.

u = 0 m/s

a = - 9.8 $\frac{m}{s^2}$ (gravity)

    v = ???

$v^2 = u^2 + 2as$           

$v^2 = 0^2 + 2(-9.8)(15)$

Do the math and we should get that $v=\pm\sqrt{294}$. 

Common sense says that since the person is falling, their velocity should be negative,

so $v = -17.14643$

Now that we have the velocity, we can solve for the acceleration of the person when they're in the net. 

SUVAT again.

s = 1.0 m. 

u = -17.14643    

v = 0 m/s (the net slows them to rest)   

a = ???

$v^2 = u^2 + 2as$         

$0^2 = (17.14643)^2 + 2*1.0*a$

 

Do math and you should get that $a = -147 m/s^2$ ,

but since they asked for deceleration (meaning negative acceleration) your final answer should be

$a = -147 m/s^2$ or $a = -150 m/s^2$ with sig figs.