Magnitude of Acceloration problem

[SunKissedGal1]

Homework Statement

a person jumps from a window 46m high. and is caught in a figherfighters net which streaches .6m. to what tenth of a m/s^2 is the persons magnitude of acceloration in the net

Homework Equations

X= - 46
a = -9.8
v= ?

The Attempt at a Solution

would the soultion be -9.8? I am really not sure what formula to use or how to solve this. PLEASE HELP!

 

[asleight]

Homework Statement

a person jumps from a window 46m high. and is caught in a figherfighters net which streaches .6m. to what tenth of a m/s^2 is the persons magnitude of acceloration in the net

Homework Equations

X= - 46
a = -9.8
v= ?

The Attempt at a Solution

would the soultion be -9.8? I am really not sure what formula to use or how to solve this. PLEASE HELP!

Is this all of the given information? Something seems to be missing.

 

[SunKissedGal1]

yeah, that's all. the hit was that there's 2 parts to the problem, falling and hitting the net, idk if that helps at all...

 

[asleight]

yeah, that's all. the hit was that there's 2 parts to the problem, falling and hitting the net, idk if that helps at all...

IDK why she's in the 10th floor jumping into a net but...

Can you please write every single word, as written, with the question clearly indicated? I have no idea what needs to be solved for.

 

[SunKissedGal1]

A person jumps from a window 46 meters high and is caught in a firefighter's net which stretches 0.6 meters. to the nearest tenth of a M/S^2, what is the magnitude of the persons acceloration in the net.

that is copy/pasted from my hw.

 

[asleight]

A person jumps from a window 46 meters high and is caught in a firefighter's net which stretches 0.6 meters. to the nearest tenth of a M/S^2, what is the magnitude of the persons acceloration in the net.

that is copy/pasted from my hw.

Okay, well, here's a stab at it:

$$v_f^2=v_i^2+2ad \rightarrow v_f=\sqrt{92g}$$

From this, we'll now solve for the second portion of motion:

$$v_f^2=v_i^2+2ad \rightarrow -\sqrt{92g}=\sqrt{1.2a}$$...

 

[SunKissedGal1]

could u possibly show me the variables u used to substitute into the formula?
im kinda having a hard time with this, and I am not exacly following what your doing...

 

[asleight]

could u possibly show me the variables u used to substitute into the formula?
im kinda having a hard time with this, and I am not exacly following what your doing...

The square of your final velocity during free fall is equal to your initial velocity-squared (0m/s) plus two times the quantity of your acceleration due to gravity, g=9.81m/s/s, and your displacement, 46m. This allows you to solve for your final velocity to be, 30m/s. Then, you do the same thing for the second part. Your final velocity-squared (0m/s) is the square of your initial velocity, 902.52 plus twice your acceleration due to the net, a, and your displacement, 0.6m. Rearranging, you'll solve for a, giving you a=752m/s/s, I think. I'm rushed, calculations might be wrong.

 

[SunKissedGal1]

Thank you so much!