Trigonometry Help, question posted below @AL2006

Physics

1 answer:

Oxana [17]2 years ago

8 0

It's ' D '.

-- Period: The angle is (pi/2 · t).
From t=0 to t=4 sec, the angle changes by 2pi. So the period is 4 sec.

-- Amplitude: The biggest the cosine can ever be is 1 .
When the cosine part is 1, 'd' is 6, so the amplitude of the model is 6 .

-- Position when t=0:
When t=0, the angle in the cosine is zero.
The cosine of zero is 1 .
So when t=0, d=6 . There you are.

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Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$