A person is jumping straight up and down on a trampoline. The height of the center of mass of the person is measured every tenth of a second. It takes just over one second to complete one full bounce.

Question

(A video of this situation is displayed above.)

t (seconds) 2.90 3.00 3.10 3.20 3.30 3.40 3.50 3.60 3.70 3.80 3.90 4.00 4.10
x(t) (feet) 7.65 6.80 5.70 4.50 4.55 5.65 6.85 7.65 8.15 8.30 8.10 7.65 6.85

1.Find the average velocity (in feet/sec) of the jumper from 3.00 seconds to the time when he is at the lowest point. (If there is more than one lowest point in your data, compute the average velocity to the first "lowest point".)

(Enter answer to 2 decimal places.)

i am not sure how to do it.
i am inclined towards forming an equation using sin or cos.
the format of the equation is k+a*cos(t)
k can be calculated using the data above.
'a' which is the amplitude can be calculated using the maximum deviation from k.

BUT i am not sure ...
please help.
i am under time contraint

Answer 1

average velocity is totaldistance/totaltime

the minimum x is at t=3.50, so
x(3.00) = 6.80
x(3.50) = 4.50
∆x/∆t = -1.7/0.5 = -3.4 ft/s

If you do want to devise a function, then you see that
x(2.90) = x(3.60) = x(4.00) = 7.65
So, since x is falling at t=2.90 and again at t=4.00, the period is 1.10 seconds.

the minimum x will occur midway between t=2.90 and t=3.60, when x is at the same height, but going in opposite directions. So, we want t=3.25 seconds. That appears to make x=4.525, but clearly x=4.50 is the minimum value. So, let's say the amplitude is roughly (8.30-4.50)/2 = 1.90, with the midpoint at (8.30+4.50)/2 = 6.40

Since the maximum is at t=3.80, let's make it a cosine function shifted to that point. We end up with

x(t) = 6.40 + 1.90 cos(2π/1.10 (t-3.80))

Try plotting that. It may be a bit off since we had to approximate times and heights.