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`).

Options

A. ` 10, 20, 10`
B. ` 15, 20, 15`
C. ` 4, 15, 20`
D. ` 5, 10, 20`

Answer

B

Explanation

Time taken by the same ball to return to the hands of the juggler is `(2u)/(g) =(2xx 20)/(10) =4 s`.
So he is throwing the balls after `1 s` each. Let at some instant he throws ball mumber `4`. Before `1 s` of throwing it, he throws ball `3`. So the height of ball `3` is
`h_(3) =20 xx1-(1)/(2) 10 (1)^(2) =15 m`
Before `2 s` So the height of ball ` 2` is
`h_(2)=20 xx 2 -(1)/(2) 10 (2)^(2) =20 m`
Before `3 s`, he throws ball `1`. So the height of ball `1` is
`h_(1)=20 xx 3- (1)/(2) 10 (3)^(3) =15 m .