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** **__Solve this :__

28. A particle is thrown vertically upward, its velocity at half of the height is 10m/s, then maximum height attained by it is : (g = 10m/s^{2})

(a) 8m (b) 20m

(c) 10m (d) 16m

__Solve this :__

Dear Student,

In the present problem we use the conservation of energy.

We know,

Let the total height up to which particle raised is '*h*'. Then the velocity acquired by it at the half of the height '*h/2*' is *v'*=10m/s.

Energy at the starting point,

.....................(1)

In the middle at height '*h/2*'

................(2)

At height '*h*'

Comparing equation (2) and (3):

Regards

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