question_answer

A very large number of balls are thrown vertically upwards in quick successions in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is 5 m, the number of balls thrown per minute is: (take\[g=10m/{{s}^{2}}\])

A)  80                                         

B)  120

C)  40                                         

D)  60

Correct Answer: D

Solution :

A is the origin from where the ball is thrown. Let\[u\] be the velocity of projection of ball. \[\therefore \]  \[{{\upsilon }^{2}}={{u}^{2}}-2gh\,\,...\](upwards) At \[B,\,\,\upsilon =0\] \[u=\sqrt{2gh}=\sqrt{2\times 10\times 5}=10\,m/\sec \] Ball is projected with 10 m/sec. time taken by ball to reach max. height. \[\upsilon =u-gt\] \[0=10-10t\Rightarrow t=1\sec .\] in 1 sec. ball will reach the top and in next 1 sec it will reach the ground again. \[\therefore \]1st ball reaches B in 1 sec. \[\text{IInd}\]ball reaches B in 1 sec. When \[\text{IInd}\]is at B. 1st is at A and so on. \[\therefore \]number of ball projected\[/sec=1\] number of ball projected\[/min=60.\]