Question

A juggler keeps on moving four balls in the air throwing the balls after regular intervals. When one ball leaves his hand (speed =20 m $s^{-1}$), the position of other balls (height in m) will be (table g= 10 m $s^{-2}$)

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

Answer 1

Answer: (B)

. 15, 20, 15.

The time taken by the ball to reach its maximum height is given by the first equation of motion as,

0 = 20 - 10t = 2 sec.

So, the total time will be 4 sec i.e. juggler will touch the same ball after 4 sec only. Since he is juggling 4 balls at regular intervals, i.e. each of them is spaced at 1 sec distance interval.

Therefore the position of balls are given by the second equation of motion as follows,

$h_3$ = 20 $\times$ 1 - $\frac{1}{2}$ $\times$ 10 ${\left(1\right)}^2$ = 15 m

$h_2$ = 20 $\times$ 2 - $\frac{1}{2}$ $\times$ 10 ${\left(2\right)}^2$ = 20 m

$h_1$ = 20 $\times$ 3 - $\frac{1}{2}$ $\times$ 10 ${\left(3\right)}^2$ = 15 m