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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]
[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]
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