Question

A juggler keeps on moving four balls in the air throwing the balls after intervals. When one ball leaves his hand (speed = 20ms-1) the position of other balls (height in metre) will be (Take g = 10ms-2)

• A. 10,20,10
• B. 15,20,15
• C. 5,15,20
• D. 5,10,20

Answer 1

Answer: (B)

15,20,15

Answer 2

The time taken by a small ball to return to the juggler's hands = $\frac{2 v}{g}=\frac{2 \times 20}{10}$ = 4s.
Hence, the juggler is throwing balls after 1 second.

Let us take an instant where he is throwing ball number 4.

He throws the third ball 1 second before the 4th one. Therefore, the height of ball 3 is

h = ut - $\frac {1}{2}$ gt2

$h_{3}=20 \times 1-\frac{1}{2} 10(1)^{2}=15 \,m$

He throws the ball 2 before 2s. So the height of ball 2 is
$h_{2}=20 \times 2-\frac{1}{2} 10(2)^{2}=20\, m$

He throws the ball 1 before 3s. So the height of ball 1 is
$h_{1}=20 \times 3-\frac{1}{2} 10(3)^{2}=15\, m$

Hence, h1 is 15m.