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farnworth
- 2
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H(t) = t^3-6t^2+5t+30 this is a yo yo 30 inches above ground at t =0, at 4 secs it is 18 inches above ground. Please tell me how these figures are derived; t^3,6t^2, 5t; I realize the 30 is initial position. I am 81 but very curious. Thank you.