A man throws balls with the same speed vertic | vedclass

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(D) The interval between consecutive throws is $\Delta t = 2\,s$.For more than two balls to be in the air simultaneously,the time of flight $(T)$ of t

Vedclass · Accepted Answer

More than $19.6\,m/s$

Vedclass · Answer

(D) The interval between consecutive throws is $\Delta t = 2\,s$.For more than two balls to be in the air simultaneously,the time of flight $(T)$ of the first ball must be greater than the time taken to throw the third ball.The first ball is thrown at $t = 0$.The second ball is thrown at $t = 2\,s$.The third ball is thrown at $t = 4\,s$.For the first ball to still be in the air when the third ball is thrown,its time of flight must satisfy $T > 4\,s$.Since the time of flight $T = \frac{2u}{g}$,we have:$\frac{2u}{9.8} > 4$$2u > 39.2$$u > 19.6\,m/s$.Therefore,the speed of the throw must be greater than $19.6\,m/s$.

$A$ 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 time? (Given $g = 9.8\,m/s^2$)

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