A person throws vertically up n balls per second with the same velocity. He throws a ball whenever the previous one is at its highest point. The height to which the balls rise is

(a) g/n^{2}

(b) 2gn

(c) g/2n^{2}

(d) 2gn^{2}

Solution:

We know v=u-gt. v is zero at the highest point. Time t taken by one ball to reach maximum height is 1/n.

Hence, we have

u=gt

or, u=g/n……. (i)

now, from the relation v^{2}=u^{2}-2gh. Again, v=0

u^{2}=2gh……. (ii)

Putting the value of u from equation (i) in above relation (ii), we get

h=g/2n^{2}.

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