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4 years ago

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A sack of potatoes weighing 16.0-kg falls from a very tall building. At a certain point at the motion downwards, its measured ac

celeration is 4.6 m/s2 and its velocity is 54.2 m/s. Assuming that the magnitude of the drag force due to air resistance is proportional to the square of its speed, What is its terminal speed?

1 answer:

Vitek1552 [10]4 years ago

7 0

Answer:

The terminal speed is 74.833 m/s

Explanation:

The drag force is equal to square of speed:

Fdrag = k*v²

According Newton`s law:

Fnet = m*a

m*g - k*v² = m*a

$k=\frac{m(g-a)}{v^{2} }$

$k=\frac{16*(9.8-4.6)}{54.2^{2} } =0.028$

If terminal speed, the net force is zero.

$kv_{t} ^{2} =mg\\v_{t} =\sqrt{\frac{mg}{k} } =\sqrt{\frac{16*9.8}{0.028} } =74.833m/s$

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Answer: $0.8\ mm$

Explanation:

Given

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$\Delta l=\dfrac{Pl}{AE}$

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