$\"\"$

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The stone is dropped from the top of the tower.

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The height of the tower from the ground is $\"\"$ m.

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(a)

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Find the distance of the stone at time $\"\"$ .

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At the initial position, $\"\"$ , $\"\"$ and $\"\"$ .

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The acceleration of the object is $\"\"$ , where $\"\"$ is the velocity.

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Hence $\"\"$

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The antiderivative of $\"\"$ is $\"\"$ .

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Substitute $\"\"$ and $\"\"$ in $\"\"$ .

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$\"\"$

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Therefore the velocity of the stone is $\"\"$ .

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The velocity of the object is $\"\"$ , where $\"\"$ is the distance.

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Hence $\"\"$ .

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The antiderivative of $\"\"$ is

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$\"\"$

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Substitute $\"\"$ and $\"\"$ in $\"\"$ .

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$\"\"$

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Therefore the distance of the stone at time $\"\"$ is $\"\"$ .

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$\"\"$

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(b)

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The stone reaches the ground when $\"\"$ .

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$\"\"$

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The stone takes $\"\"$ s to reach the ground.

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$\"\"$

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(c)

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The velocity of the stone is $\"\"$ .

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At time $\"\"$ , the stone reaches the ground.

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$\"\"$

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Here negative sign indicates that the direction of the stone (downward).

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The velocity of the stone strikes the ground is $\"\"$ m/s.

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$\"\"$

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(d)

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The velocity of the stone is $\"\"$ .

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Here $\"\"$ , since the stone is thrown downward.

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$\"\"$

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The velocity of the stone is $\"\"$ .

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The velocity of the object is $\"\"$ , where $\"\"$ is the distance.

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Hence $\"\"$ .

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The antiderivative of $\"\"$ is

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$\"\"$

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Substitute $\"\"$ and $\"\"$ in $\"\"$ .

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$\"\"$

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Therefore the distance of the stone at time $\"\"$ is $\"\"$ .

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The stone reaches the ground when $\"\"$ .

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$\"\"$

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$\"\"$

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$\"\"$

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The stone takes $\"\"$ s to reach the ground with a velocity $\"\"$ m/s.

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$\"\"$

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(a) The distance of the stone at time $\"\"$ is $\"\"$ .

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(b) The stone takes $\"\"$ s to reach the ground.

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(c) The velocity of the stone strikes the ground is $\"\"$ m/s.

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(d) The stone takes $\"\"$ s to reach the ground with a velocity $\"\"$ m/s.