Question

A person jumps from the roof of a house 3.1-m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70 m. If the mass of his torso (excluding legs) is 42kg .
Part A
Find his velocity just before his feet strike the ground.
Express your answer using two significant figures. Enter positive value if the velocity is upward and negative value if the velocity is downward.

Answer 1

Part B
Find the average force exerted on his torso by his legs during deceleration.
Express your answer using two significant figures. Enter positive value if the force is upward and negative value if the force is downward.

Answer 2

i got an answer of -6.9 m/s but it wrong

Answer 3

that was for part a and part b i got 33.6 m/s2

Answer 4

to get the answer for A simply use conservation of mechanical energy
mgh=mv^2/2
v=sqrt(2gh) pointing down

Answer 5

ahh thank you

Answer 6

i got 7.79, how would i do part b

Answer 7

i got an answer of 1820.523 N i dont know if i am correct

Answer 8

One of the one-dimensional motion equations is relevant here:\[v^2=v_0^2+2ax\]Doing some algebra, plugging in numbers, and using Newton's second law, you'll find the answer you got:

+1821N.

Remember, though, that your reported velocity should be negative, since it's in the "down" direction.

Also, report using the correct number of significant figures.

Answer 9

i have a question would i use the x of 0.70 or 3.1 m

Answer 10

The equation should really be written like this:\[v_f^2=v_i^2+2a(\Delta x)=v_i^2+2a(x_f-x_i)\]The v_f and the x_f are corresponding values. Whatever the velocity you used for v_f should have a corresponding position, x_f. The same goes for v_i and x_i.

Since you're looking at the time between him hitting the ground and completely stopping moving, you are looking at 0.70m (x_i) and 0m (x_f) as well as -7.79m/s (v_i) and 0m/s (v_f).

You just need to be sure to use positions and velocities that correspond with each other if that makes sense... Maybe my explanation isn't the best.

Answer 11

no i do get it, thank you so much =)

Answer 12

and if its in 2 sig figs it would be 1.8 x 10^3

Answer 13

That's right (:.

Answer 14

it saying my answer is wrong

Answer 15

i dont know what i did wrong

Answer 16

Did you include units?

Answer 17

yes i put N

Answer 18

Was your velocity value correct?

Answer 19

yes i used 7.8

Answer 20

well -7.8

Answer 21

I'm afraid I don't know. The best I can say is to try reporting it as 1800N, but otherwise either we're wrong or the program is just not accepting that answer for whatever reason.

Answer 22

@Luigi0210

Answer 23

Physics?

Answer 24

yes

Answer 25

I have never taken physics in my life: @agent0smith

Answer 26

oki thats fine thank u

Answer 27

haha i solved this problem for my AP physics class a few days ago.

Answer 28

i tried so many times to figure this out but i got it wrong

Answer 29

Oh in my book it was 3.9m tall... are you sure about the 3.1m? Its the Giancoli Physics book, 6th Ed.

Answer 30

All other numbers are the same, except 3.9 m not 3.1 m.

Answer 31

i have the 7th edition

Answer 32

and this is homework we have to submit online

Answer 33

Oh, well... then idk, i don't see why 1.8x10^3 N is wrong :/

Answer 34

hmm how did u do your problem out?

Answer 35

Same way you and @Xishem did, using the same equations and F=ma.

Answer 36

idk what distance to use?

Answer 37

would it be 0.70 or 3.1 for part b

Answer 38

for part b, you know the velocity, and that his torso decelerates over 0.7m. Use it to find accel.

Answer 39

i did that then i plugged that into f=ma and got the force

Answer 40

Yep, idk why 1800N is wrong. I got that.

Answer 41

You could always try... 2.3x10^3 N, i thinkk that was the answer i got from my book's question (it's an even question and the answers only show odd questions so i can't actually check it). But i can't imagine why they'd update the velocity answer but not the force answer

Answer 42

okay thank you

Answer 43

can you help me out on another problem?

Answer 44

i can post it in another section

Answer 45

ok

Answer 46

@Xishem we forgot that the NET force is what decelerates the torso, but this isn't the average force exerted ON his torso BY his legs during deceleration.