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Marjorie drops an object from a height of [tex]\( h \)[/tex] meters, and it hits the ground with a velocity of [tex]\( v \, \text{m/s} \)[/tex], as given by the function [tex]\( v = \sqrt{19.6 h} \)[/tex].

If the velocity of the object is [tex]\( 71.6 \, \text{m/s} \)[/tex], what was the height of the object before it was dropped? Round your answer to the nearest tenth.

Type your numerical answer below.

[tex]\[ h = \][/tex] meters


Sagot :

To determine the height [tex]\( h \)[/tex] from which Marjorie dropped the object, given that it hits the ground with a velocity of [tex]\( 71.6 \, \text{m/s} \)[/tex], we start with the given function:

[tex]\[ v = \sqrt{19.6h} \][/tex]

We need to isolate [tex]\( h \)[/tex]. Here are the steps to solve for [tex]\( h \)[/tex]:

1. Substitute the given velocity value [tex]\( v = 71.6 \, \text{m/s} \)[/tex] into the equation:
[tex]\[ 71.6 = \sqrt{19.6h} \][/tex]

2. To eliminate the square root, square both sides of the equation:
[tex]\[ (71.6)^2 = (\sqrt{19.6h})^2 \][/tex]

3. Simplify the left side:
[tex]\[ 5130.56 = 19.6h \][/tex]

4. Solve for [tex]\( h \)[/tex] by dividing both sides by 19.6:
[tex]\[ h = \frac{5130.56}{19.6} \][/tex]

5. Perform the division:
[tex]\[ h = 261.6 \, \text{meters} \][/tex]

Thus, the height from which the object was dropped is:

[tex]\[ h = 261.6 \, \text{meters} \][/tex]