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------------------------------------------------ To defend the castle, a rock is thrown downwards from the top of the wall with an initial speed of 7.40 m/s. What is the rock’s speed as it passes the height of 1.55 m from the ground?

To find the speed of the rock as it passes the height of 1.55 m from the ground, we can use the principle of conservation of energy. At the top of the wall, the rock has potential energy, which gets converted into kinetic energy as the rock falls. Using the equation for potential energy and kinetic energy, we can find the speed of the rock at 1.55 m above the ground.

Using the equation for potential energy, E = mgh, where m is the mass of the rock, g is the acceleration due to gravity, and h is the height, we have:

mgh = (1/2)mv^2

Canceling out m and rearranging the equation, we get:

v = sqrt(2gh)

Substituting the known values, we have:

v = sqrt(2 * 9.8 * 1.55) = 6.98 m/s

Therefore, the speed of the rock as it passes the height of 1.55 m from the ground is approximately 6.98 m/s.

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