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Reil [10]
3 years ago
10

A ball is thrown straight up in the air. When will its kinetic energy be the least before it is caught? A. at the start of its f

light. . B. halfway to the top of its flight. . C. at the end of its flight. . D. at the top of its flight
Physics
2 answers:
svet-max [94.6K]3 years ago
4 0

The correct answer to the question is D). At the top of its flight.

EXPLANATION:

When a  ball is thrown into air, the only force except air resistance that affects the motion of the ball is the force of gravity which pulls the body in vertically  downward direction.

When the ball was thrown, it has kinetic energy only. As the ball went up, its kinetic energy is converted into potential energy.

Due to the gravity,the velocity of the ball is decreasing each time. At a certain time, the ball will reach at its maximum height where its velocity will be completely zero. At that time whole of its kinetic energy must have converted into potential energy due to the law of conservation of energy.

Hence, the kinetic energy is least at the top of its flight.

vlabodo [156]3 years ago
3 0

Your answer is

D.) At The Top Of Its Flight

because once something gets to the top, or let's say the peak of its journey, it has gathered the most energy


Hope this Helps! :)

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Answer:

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a 15 kg tv sots on a shelf at a height of 0.3 m. how much gravitational potential energy is added to the television when it is l
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Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa
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The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

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Given that,

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E_{1}=\dfrac{kq}{r'^2}

Put the value into the formula

E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}

E_{1}=52200=52.2\times10^{3}\ N/C

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Using formula of electric field

E_{1}=\dfrac{kq}{r^2}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}

E_{2}=104400=104.4\times10^{3}\ N/C

We need to calculate the horizontal electric field

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E_{x}=52.2\times10^{3}\times\cos45

E_{x}=36910.97=36.9\times10^{3}\ N/C

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E_{net}=\sqrt{E_{x}^2+E_{y}^2}

Put the value into the formula

E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}

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\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})

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Learn More:

For more on P- and S-- waves check out;

brainly.com/question/11915788

brainly.com/question/11334414

brainly.com/question/2530620

#LearnWithBrainly

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