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3 years ago

What would be the speed of an object just before hitting the ground if dropped 91.5 meters

Physics

1 answer:

Aleks04 [339]3 years ago

5 0

Answer:

about 42.35 m/s

Explanation:

Use the equation for accelerated motion (g), and with zero initial velocity that doesn't include time:

$v_f^2=v_i^2+2\,a\,\Delta x$

which for our case would reduce to:

$v_f^2=v_i^2+2\,a\,\Delta x\\v_f^2=0+2\,9.8\,(91.5)\\v_f^2= 1793.4\\v_f=\sqrt{1793.4} \\v_f \approx 42.35$

then the velocity just before hitting would be about 42.35 m/s

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Answer:

Part a)

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Explanation:

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part b)

$k = \frac{\omega}{v}$

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so we have

$k = \frac{579.3}{5.69}$

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Part c)

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