Answer
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Hint:First of all to calculate the potential energy we know the formula \[P.E = mgh\] where these are mass, acceleration due to gravity and height of ball at that point and then kinetic energy using \[K.E = \dfrac{{m{v^2}}}{2}\],here velocity is considered as present velocity of particle.
Complete step-by-step solution:
We know potential energy at half way will be half of potential energy at top and total energy remains conserved all over the path so we know initially total energy will be in the form of kinetic energy as height is zero so total energy is in the form of kinetic energy, therefore calculating kinetic initially,
\[T.E = K.E\]
\[T.E = \dfrac{{0.1 \times {{\left( {20} \right)}^2}}}{2}\]
\[T.E = \dfrac{{0.1 \times 400}}{2}\]
\[T.E = 20\]
And at top total energy will be potential energy and the same as calculated above therefore at mid potential energy will be half of total energy.
\[P.E = \dfrac{{T.E}}{2}\]
\[P.E = 10\]
And as we know total energy remains conserver all over the path and as we have calculated the potential energy above now kinetic energy will be easily calculate by:
\[K.{E_{mid}} = T.E - P.{E_{ha\operatorname{l} f}}\]
\[K.{E_{mid}} = 20 - 10\]
\[K.{E_{mid}} = 10\]
Therefore the correct option is B.
Additional Information:In the event that no powers (other than gravity) follow up on the ball during its excursion up and withdraw, at that point the ball's all out mechanical vitality (dynamic + potential) won't change. Be that as it may, the vitality of the ball can change structures. At first, the ball has a ton of dynamic vitality, since you've given it a really huge upward speed. For this situation, the complete vitality is saved in light of the fact that it doesn't change. Be that as it may, as the ball moves upward, it eases back down as its underlying active vitality is changed into expected vitality. In the end, the entirety of the underlying active vitality will become likely vitality, and the ball will stop immediately. The ball has at last arrived at its most elevated point. After this, it'll pivot and fall back to the ground as all the potential vitality it had at the most noteworthy point is changed once more into active vitality as it tumbles to the ground.
Note:- In this question there was one more method to calculate the potential energy at half without using the concept of total energy is by calculating the total height as initial velocity is given and then calculating the velocity at the mid but it will be a longer method.
Complete step-by-step solution:
We know potential energy at half way will be half of potential energy at top and total energy remains conserved all over the path so we know initially total energy will be in the form of kinetic energy as height is zero so total energy is in the form of kinetic energy, therefore calculating kinetic initially,
\[T.E = K.E\]
\[T.E = \dfrac{{0.1 \times {{\left( {20} \right)}^2}}}{2}\]
\[T.E = \dfrac{{0.1 \times 400}}{2}\]
\[T.E = 20\]
And at top total energy will be potential energy and the same as calculated above therefore at mid potential energy will be half of total energy.
\[P.E = \dfrac{{T.E}}{2}\]
\[P.E = 10\]
And as we know total energy remains conserver all over the path and as we have calculated the potential energy above now kinetic energy will be easily calculate by:
\[K.{E_{mid}} = T.E - P.{E_{ha\operatorname{l} f}}\]
\[K.{E_{mid}} = 20 - 10\]
\[K.{E_{mid}} = 10\]
Therefore the correct option is B.
Additional Information:In the event that no powers (other than gravity) follow up on the ball during its excursion up and withdraw, at that point the ball's all out mechanical vitality (dynamic + potential) won't change. Be that as it may, the vitality of the ball can change structures. At first, the ball has a ton of dynamic vitality, since you've given it a really huge upward speed. For this situation, the complete vitality is saved in light of the fact that it doesn't change. Be that as it may, as the ball moves upward, it eases back down as its underlying active vitality is changed into expected vitality. In the end, the entirety of the underlying active vitality will become likely vitality, and the ball will stop immediately. The ball has at last arrived at its most elevated point. After this, it'll pivot and fall back to the ground as all the potential vitality it had at the most noteworthy point is changed once more into active vitality as it tumbles to the ground.
Note:- In this question there was one more method to calculate the potential energy at half without using the concept of total energy is by calculating the total height as initial velocity is given and then calculating the velocity at the mid but it will be a longer method.
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