Suppose that an object is dropped from a height of $$ h $$ meters and hits the ground with a velocity of $$ v $$ meters per second. Then $$ v=\sqrt{19.6 h} $$. If an object is dropped from o height of 22.1 meters, with what velocity does it hit the ground? Round your answer to the nearest tenth.

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Given that the object is dropped from a height of $$ h $$ meters and hits the ground with a velocity of $$ v $$ meters per second, the relationship between the height and the velocity is given by the equation $$ v = \sqrt{19.6h} $$.

To find the velocity with which the object hits the ground when dropped from a height of 22.1 meters, we can substitute $$ h = 22.1 $$ into the equation and solve for $$ v $$.

Substitute $$ h = 22.1 $$ into the equation $$ v = \sqrt{19.6h} $$:
$$ v = \sqrt{19.6 \times 22.1} $$

Now, we can calculate the value of $$ v $$ by simplifying the expression inside the square root and then taking the square root of the result.
Calculate the value by following steps:
- step0: Calculate:
  $$\sqrt{19.6\times 22.1}$$
- step1: Multiply the numbers:
  $$\sqrt{433.16}$$
- step2: Rewrite the expression:
  $$\sqrt{\frac{10829}{25}}$$
- step3: Use the properties of radicals:
  $$\frac{\sqrt{10829}}{\sqrt{25}}$$
- step4: Simplify the expression:
  $$\frac{7\sqrt{221}}{5}$$
The velocity with which the object hits the ground when dropped from a height of 22.1 meters is approximately 20.8 meters per second.

Rounding this value to the nearest tenth, the velocity is 20.8 meters per second.
The object hits the ground with a velocity of approximately 20.8 meters per second.

Suppose that an object is dropped from a height of \( h \) meters and hits the ground with a velocity of \( v \) meters per second. Then \( v=\sqrt{19.6 h} \). If an object is dropped from o height of 22.1 meters, with what velocity does it hit the ground? Round your answer to the nearest tenth.

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