The time taken by ball to reach maximum height

v = u − gT

At maximum height, final speed is zero ie, v = 0

so, u = gT


\( T = frac{u}{g}\)

In 2s,

= u−2 * 9.8 − 19.6ms−1

If a man throws the ball with a velocity of 19.6ms−1, then after 2s, it will reach the maximum height.

When he throws the 2nd ball, the 1st will be at the top.

When he throws the third ball, 1st will come to the ground and 2nd will be at the top.

Therefore, only two balls are in the air.

If he wants to keep more than two balls in the air, he should throw the ball with a speed greater than 19.6ms−1

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