5h+3 - j; use h = 6, and j = 1
2 answers:
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Answer:
1.63 cm (across the centerline from release)
Step-by-step explanation:
If we assume time starts counting when we release the weight from its fully-extended downward position, then the position at 1.15 seconds can be found from ...
h(t) = -7cos(2πt/4)
h(1.15) = -7cos(π·1.15/2) = -7(-0.233445) ≈ 1.63412 . . . cm
That is, 1.15 seconds after the weight is released from below the resting position, it will be 1.63 cm above the resting position.
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If it is released from <em>above</em> the resting position, it will be 1.63 cm <em>below</em> the resting position at t=1.15 seconds.
