Question

A juggler keeps on moving four balls in air by throwing the balls after regular intervals. He throws the ball with a speed of 60 m/s. When one ball leaves his hand the position of other balls from his hand will be

1. 135 m, 180 m, 225 m

2. 135 m, 180 m, 135 m

3. 180 m, 225 m, 270 m

4. 120 m, 150 m, 180 m​

Answer 1

Answer:

Option: 4) 120m, 150m, 180m

Explanation:

might be wrong coz I'm not that good in math

Answer 2

Time taken by the same ball to return to the hands of the juggler(t) $\\ = 2\frac{u}{g}$

Where. u= initial velocity.

 $\\ t=\frac{2u}{g}\\\\ = \frac{2 \times 60}{10}\\\\ = 12s$

So, he is throwing the ball after each 3 s.

Let at some moment, he is throwing ball number 4.

Before 3 s of it, he would throw ball 3.

So, the height of ball 3=

$h_{3} = 60\times3 - \frac{1}{2} \times 10\times 3^{2} = 135m$

Before 6 s of it, he would throw ball 2.

So, the height of ball 2=

 $h_{2} = 60\times6 - \frac{1}{2} \times 10\times 6^{2} = 180m$

Before 9 s of it, he would throw ball 1.

So, the height of ball 1=

 $h_{1} = 60\times9 - \frac{1}{2} \times 10\times 9^{2} = 135m$

Hence, the correct option is (2) 135 m, 180 m, 135 m