A man throws balls with the same speed vertically upwards one after the other at an interval of $2$ s What should be the speed of the throw so that more than two balls are in the sky at any rime?

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Solution

When you throw the ball upwards the final velocity becomes zero, therefore,

$v = u + a t$

$0 = u - g t$

$u = 10 \times 2 = 20 m / s$

Hence, when you throw a ball with minimum $20 m / s$ in a time interval of two seconds two balls can be in the air simultaneously...

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A man throws balls with the same speed vertically upwards one after the other at an interval of 2 s What should be the speed of the throw so that more than two balls are in the sky at any rime?

When you throw the ball upwards the final velocity becomes zero, therefore, v=u+at 0=u−gt u=10×2=20m/s Hence, when you throw a ball with minimum 20m/s in a time interval of two seconds two balls can be in the air simultaneously...

A man throws balls with the same speed vertically upwards one after the other at an interval of $2$ s What should be the speed of the throw so that more than two balls are in the sky at any rime?

When you throw the ball upwards the final velocity becomes zero, therefore,

$v = u + a t$

$0 = u - g t$

$u = 10 \times 2 = 20 m / s$

Hence, when you throw a ball with minimum $20 m / s$ in a time interval of two seconds two balls can be in the air simultaneously...