1.

A juggler keeps on moving four balls in the air throwing the balls after regular intervals. When one ball leaves his hand `("speed" = 20 ms^-1)` the positions of other balls (height in m) `("Take" g = 10 ms^-2)`.A. `10,20,10`B. `15, 20,15`C. `5, 15,20`D. `5,10,20`

Answer» Correct Answer - B
Time taken by same ball to return to the hands of juggler `= (2 u)/(g) = (2 xx 20)/(10) = 4 s`. So he is throwing the balls after each `1 s`. Let at some instant he is throwing ball number `4`. Before `1 s` of it he throws ball `3`. So height of ball `3` :
`h_3 = 20 xx 1 - (1)/(2) 10(1)^2 = 15 m`
Before `2 s`, he throws ball `2`. So height of ball `2` :
`h_2 = 20 xx 2 - (1)/(2) 10(2)^2 = 20 m`
Before `3 s`, he throws ball `1`. So height of ball `1` :
`h_1 = 20 xx 3 - (1)/(2) 10(2)^2 = 20 m`.


Discussion

No Comment Found

Related InterviewSolutions