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Hitman42 [59]
3 years ago
12

A rock falls off a 15 meter cliff. How fast is the rock traveling vertically two seconds later?

Physics
1 answer:
oksian1 [2.3K]3 years ago
8 0
The acceleration of gravity on Earth is 9.8 m/s².  That means that
an object falling under the influence of gravity will move 9.8 m/s
faster than it was moving a second earlier.

Falling from rest, it will be moving 9.8 m/s after the first second,
and 19.6 m/s after the 2nd second.

The height from which it fell doesn't matter.
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One situation in which force is created is when an object is moving and a force is created to stop that movement. Second situation is when an object is moving circularly and a force is created to move it towards the middle of the circle. The third situation is when a force is created that goes in the same direction as an object that is in movement.
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A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calcul
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Answer:

Part i)

h = 5.44 m

Part ii)

h = 3.16 m

Explanation:

Part i)

Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy

So we have

\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh

here we know that for spherical shell and pure rolling conditions

v = R \omega

I = \frac{2}{3}mR^2

\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = mgh

\frac{5}{6}mv^2 = mgh

h = \frac{5v^2}{6g}

h = \frac{5(8^2)}{6(9.81)} = 5.44 m

Part b)

If ball is not rolling and just sliding over the hill then in that case

\frac{1}{2}mv^2 = mgh

h = \frac{v^2}{2g}

h = \frac{8^2}{2(9.81)} = 3.16 m

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