1.

a boy throws n balls per second at regular time intervals when the first ball reaches the maximum height he throws the second one vertically up the maximum height reached by each ball is​

Answer»

ong>Answer:

Time GAP between two successive throws is t=

n

1

so it is the time taken by one ball to reach the top point.

In other words it is the time taken to make the velocity zero so we can calculate the velocity of projection.

If velocity of projection is u then final velocity at top point(comes to rest) i.e v=0=u−gt=u−g(

n

1

)

so u=g/n

Now we KNOW that for projection velocity, u the maximum height is H=

2g

u

2

=

n

2

×2g

g

2

=

2n

2

g

Option B is correct.


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