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3 years ago

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A stone with a mass m is dropped from an airplane that has a horizontal velocity v at a height h above a lake. If air resistance

is neglected, the horizontal distance R from the point on the lake directly below the point of release to the point where the stone strikes the water is given by which formula? A) R = v(2h/g)^2 B) R = (1/2)gt^2 C) R = 2mv squareroot (2h/g) D) R = v squareroot (2h/g) E) None of these is correct.

1 answer:

seropon [69]3 years ago

4 0

Answer: Option B. R = (1/2)gt^2

Explanation:

S = R (horizontal distance)

V^2 = 2gS

V^2 = 2gR

R = V^2 / 2g

But V = gt

R = (gt)^2 / 2g

R = (g^2 x t^2) / 2g

R = gt^2 / 2

But t^2 = 2h/g

R = ( g x 2h/g) / 2

R = h

But h = (1/2)gt^2

R = h = (1/2)gt^2

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Where:

$F$ - Force, measured in newtons.

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From (Eq. 1) we get the following relationship and clear the final force within:

$F_{A}\cdot r_{A} = F_{B}\cdot r_{B}$

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$F_{A}$ , $F_{B}$ - Initial and final forces, measured in newtons.

$r_{A}$ , $r_{B}$ - Initial and final distances, measured in meters.

If we know that $F_{A} = 5\,N$ , $r_{A} = 0.9\,m$ and $r_{B} = 0.35\,m$ , then final force is:

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2 years ago

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3 years ago

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<u>Given:</u>

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<u>Assume:</u>

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