Question

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 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.)

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

Answer 1

The impulse received from the floor is, 0.800 0 0.800 -i (92001-11500t*)dt 9200r2 1150023 0800 2 (4600) (0.800-(38333 -(4600) (0.800) - (3833.33) (0.800) 981 N-S The speed when the woman reached the floor is, 2, V2gh 2(9.8m/s)(0.790m) -3.934 m/s

The net impulse that acts on the women is, net -98 1 N.s_ (68 kg)(9.8 m/s)(0.8s) 447.88 N-s From the definition of impulse, it is equal to change in momentum m(v+v) Use above equation to solve for speed that the woman leaves from the floor m(v+v)-447.88N-s 447.88N.s 447.88N.s - 3.934 m/s 68 kg -2.652 m/s Use the following kinematic equation to solve for height 2g ,. 02 -(2.652 m/s 2-9.8 m/s h 0.3 58 m