Starting from rest, a 68.0 kg woman jumps down to the floor from a height of 0.790 m, and immediately jumps back up into the air. While she is in contact with the ground during the time interval0 < t < 0.800 s,the force she exerts on the floor can be modeled using the functionF = 9,200t − 11,500t2where F is in newtons and t is in seconds.(a) What impulse (in N · s) did the woman receive from the floor? (Enter the magnitude. Round your answer to at least three significant figures.)in N · s(b) With what speed (in m/s) did she reach the floor? (Round your answer to at least three significant figures.)in m/s(c) With what speed (in m/s) did she leave it? (Round your answer to at least three significant figures.)in m/s(d) To what height (in m) did she jump upon leaving the floor?in m

The force of the woman on the floor is given and by action and rection the floor exerts on the woman a force of equal magnitude, but opposite direction

I = ∫ (9200 t - 11500 t2) dt

I = 9200 t² / 2 - 11500 t³ / 3

We evaluate between the lower limit t = 0 and upper limit t = 0.800 s

I = 9200 (0.8² -0) - 11500 (0.8³ -0)

I = 5888 -5888

I = 0 N s

Directed from the floor to the woman

b) For this part we use kinematics

v² = v₀² - 2g y

v = √ (0 - 2 9.8 (-0.79))

v = 3.935 m / s

The speed direction is down

c) for this we use the relationship between momentum and the amount of movement

I = ΔP

I = m vf - m v₀

vf = (I + m v₀) / m

This is the impulse of women on the floor

vf = ( 0 + 68 (3.935)) / 68

vf = 3.935 m / s

d) let's use kinematics

v₂ = v₀² - 2gy

0 = v₀² - 2gy

y = v₀² / 2g

y = 3.935²/2 9.8

y = 0.790 m