A person throws balls into air after every second. The next ball is thrown when the velocity of the first ball is zero. How high do the ball rise above his hand?

2 m

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5 m

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8 m

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10 m

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Solution

The correct option is B $5 m$
According to question, second ball is thrown $1 s$ after the first ball and at this instance first ball is at its maximum height. This is because at maximum height, velocity of the ball is $0$ .
$\Rightarrow$ Time taken by the ball to reach maximum height is $1 s$ .
$\Rightarrow$ Height $= \frac{1}{2} g t^{2} = 5 m$

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A person throws balls into air after every second. The next ball is thrown when the velocity of the first ball is zero. How high do the ball rise above his hand?

The correct option is B 5 mAccording to question, second ball is thrown 1 s after the first ball and at this instance first ball is at its maximum height. This is because at maximum height, velocity of the ball is 0. ⇒ Time taken by the ball to reach maximum height is 1 s. ⇒ Height =12gt2=5 m