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Juli2301 [7.4K]

3 years ago

8

You throw a rock straight up into the air with a speed of 14.2 m/s. how long does it take the rock to reach its highest point?

1 answer:

slega [8]3 years ago

4 0

The acceleration of gravity on or near the Earth's surface is 9.8 m/s² downward.
Is that right ? I don't hear any objection, so I'll assume that it is.

That means that during every second that gravity is the only force on an object,
the object either gains 9.8m/s of downward speed, or it loses 9.8m/s of upward
speed. (The same thing.)

If the rock starts out going up at 14.2 m/s, and loses 9.8 m/s of upward speed
every second, it runs out of upward gas in (14.2/9.8) = <em>1.449 seconds</em> (rounded)

At that point, since it has no more upward speed, it can't go any higher. Right ?

(crickets . . .)

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Answer:

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substituting this into the equation

$KE = \frac{ 1}{2} *\frac{GMm}{r}$

Now the gravitational force of the planet is mathematically represented as

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$KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r$

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