How can I solve this?

Question

Starting from rest, a 62.0 kg woman jumps down to the floor from a height of 0.640 m, and immediately jumps back up into the air. While she is in contact with the ground during the time interval 0 < t < 0.800 s, the force she exerts on the floor can be modeled using the function F = 9,200t − 11,500t² where 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.)

______ N · s

(b) With what speed (in m/s) did she reach the floor? (Round your answer to at least three significant figures.)

 ______ m/s

(c) With what speed (in m/s) did she leave it? (Round your answer to at least three significant figures.)

 ______ m/s

(d) To what height (in m) did she jump upon leaving the floor?

______ m

Answer 1

Let

m = 62kg be the mass of the woman,

g = 9.8 m/s² be gravitational acceleration,

h₀ = 0.64m be the starting height,

h₁ be the final height (to be determined in part (d)),

v₀ be the velocity with which she reached the floor (to be determined in part (b)),

v₁ be the velocity with which she left the floor (to be determined in part (c)).

First consider the relationship between a height h and velocity v in parts (b) and (c).

In both parts there is a conservation of energy:

Kinetic energy due to velocity v, namely mv²/2,

equals potential energy at height h, namely hmg.

Thus

mv²/2 = hmg

v² = 2hg (1)

(a) By definition of impulse, it is the integral of force over a time interval.

The indefinite integral of F is 9200t²/2 - 11500t³/3 = t²(4600 - 3833t) + C

The definite integral of F from 0 to 0.8s is

J = 0.8²(4600 - 3833 . 0.8) = 981 kgm/s

(b) Using identity (1)

v₀² = 2h₀g

|v₀| = sqrt(2h₀g)

Substituting actual numbers:

|v₀| = sqrt(2 . 0.64m . 9.8 m/s²) = 3.54 m/s

v₀ is a vector pointing downward, therefore we chose the negative square root:

v₀ = -3.54 m/s

(c) We use the fact that impulse equals change in momentum

J = mv₁ - mv₀

v₁ = J/m + v₀

Substituting actual numbers

v₁ = 981kgm/s / 62kg - 3.54m/s = 12.28 m/s

(d) Using identity (1)

v₁² = 2h₁g

h₁ = v₁²/2g

Substituting actual numbers

h₁ = (12.28m/s)² / (2 . 9.8 m/s²) = 7.69 m