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

a) I = 0 N s,  b)  v = -3.935 m / s, c) vf = 3.935 m / s,   d)   y = 0.790 m

Explanation:

a) Let's start by defining the upward direction (+ y) as positive. For this part of the exercise we must use the momentum relationship

          I = ∫F dt

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