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GREYUIT [131]
3 years ago
8

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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Here,

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The direction of the velocity will be given by the tangent of the components, then

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\theta = tan^{-1} (\frac{21.77}{14.89})

\theta = 55.59\°

The magnitude is given vectorially as,

|V| = \sqrt{v_x^2+v_y^2}

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