Question

A force platform is a tool used to analyze the performance of athletes measuring the vertical force that the athlete exerts on the ground as a function of time. Starting from rest, a 67.0-kg athlete jumps down onto the platform from a height of 0.720 m. While she is in contact with the platform during the time interval 0 < t < 0.800 s, the force she exerts on it is described by the functionF = 9 200t − 11 500t2where F is in newtons and t is in seconds.(a) What impulse did the athlete receive from the platform?N · s up(b) With what speed did she reach the platform?m/s(c) With what speed did she leave it?(d) To what height did she jump upon leaving the platform?

Answer 1

a.
`I=981.34 N*s`

b.
`v_f=3.96 m/s`

c.
`v_(f1)=3.63m/s`

d.
`y_f=0.673m`

Step-by-step explanation:

Given:
`m=67kg`,
`h=0.720m`,
`0<t<0.80s`

a.

`I=\int\limits^(t_1)_(t_2) {F(t)} \, dt`

`F(t)=9200*t-11500t^2`

`I=\int\limits^(0.8s)_(0s){9200*t-11500*t^2} \, dt`

`I=4600*t^2-3833.3*t^3|(0.80,0)`

`I=2944-1962.66=981.35`

`I=981.34 N*s`

b.

`v_f^2=v_i^2+a*y'`

Starting from the rest

`v_f^2=0+2*9.8m/s^2*0.80s`

`v_f^2=15.68`

`v_f=√(15.68m^2/s^2)=3.96 m/s`

c.

`I_(total)=p_f`

`I_1-m*g*d=m*v_(f1)-m*v_f`

`981.34-67kg*9.8m/s^2*0.720=67.0kg*v_(f1)-67.0kg*(-3.96m/s)`

Solve to vf

`v_(f1)=3.63m/s`

d.

`v_f^2=v_i^2+2*a*y_f'`

`y_f'=v_i/2*a =(3.63m/s)^2/2*9.8m/s^2`

`y_f=0.673m`