Question

During a running sprint the runner’s foot is on the ground for 0.097 s and the...

During a running sprint the runner’s foot is on the ground for 0.097 s and the person is in the air for 0.118 s. The runner has a mass of 94 kg. His drag area is 0.58 m2 and air density is 1.11 kg/m3. He s running at an average velocity of 10.51 m/s over level ground.

What is the average GRFy (in units of Newtons) during the ground contact phase?

In order to maintain his average velocity, how much average GRFx must he produce during the ground contact phase?

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Answer #1

speed of the runner when he leaves the ground=acceleration due to gravity*(time in which he is in air/2)

=9.81*0.118/2

=0.57879/s

he will touch down on the ground with this speed but in opposite direction.

impulse applied=change in momentum

==>force*time=mass*(final velocity-initial velocity)

==>force*0.097=94*(0.57879-(-0.57879))

==>force=94*0.57879*2/0.097=1121.77 N

then total reaction force from the ground=weight of the runner+ reaction force during the contact

=1121.77+94*9.81=2043.91 N

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