Question

QUESTION 2 You shoot a 0.1-kg arrow straight up at 27 m/s. What is the kinetic energy when it r round your answer to the nearest ones place. half of its maximum height? Use conservation of energy and

Answer 1

First find the Kinetic energy at the bottom

$KE=1/2mv^{2}$

$KE=0.5*0.1kg*(27m/s)^{2}$

$KE=0.5*0.1*27^{2}$

$KE=36.45J$

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when the arrow reaches half of its max height, half of its kinetic energy will be converted to potential energy

$KE/2=36.45J/2$

ANSWER: $KE_{h/2}=18.225J$

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