A force platform is a tool used to analyze the performance of athletes measuring

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 57.0 kg athlete jumps down onto the platform from a height of 0.670 m. While she is in contact with the platform during the time interval 0 lessthan t lessthan 0.8 s, the force she exerts on it is described by the function below. F = (9 200 N/s)t - (11 500 N/s^2)t^2 Assume the positive y-axis points upward. What was the athlete's velocity when she reached the platform? 981.33 Ns Dimensionally incorrect. Please check the type or dimension of your unit,____j What impulse did the athlete receive from the platform? Enter a number with units. he athlete receive from gravity while in contact with the platform?____j With what velocity did she leave the platform?____J To what height did she jump upon leaving the platform?

Explanation / Answer

when the athlete is jumping from the platform:
initial speed=0 m/s
distance from ground=0.67 m
acceleration due to gravity=9.8 m/s^2
if final speed is v,
then using the formula:

final speed^2 - initial speed^2=2*acceleration*distance

==>v^2-0^2=2*9.8*0.67

==>v=3.6238 m/s

hence the athlete's velocity when she reaches the platform is 3.6238 m/s

part b:

impulse =integration of F.dt

=(9200*t*dt) - (11500*t^2*dt)

with t varying from 0 to 0.8 seconds

==>impulse=(9200*0.5*t^2)-(11500*t^3/3)

=4600*t^2-3833.33*t^3

using the limits of t from 0 to 0.8 seconds,

impulse received=981.335 N.s

c)the athlete will receive same impulse from gravity i.e. 981.335 N.s

d)as impulse applied=mass*change in velocity

if she leaves with a velocity of v1 ,

as v1 is in opposite direction of v,

impulse applied=m*(v1+v)

==>981.335=57*(v1+3.6238)

==>v1=13.5926 m/s

part e:

as maximum height reached with initial speed is given by initial speed^2/(2*g)

height reached by the athelete =v1^2/(2*g)

=13.5926^2/(2*9.8)=9.4264 m

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