Physics, asked by Brahmmi2868, 1 year ago

a juggler keeps moving four balls in air throwing the ball after regular intervals. when one leaves his hand (speed =20m/s ), find the position of other balls (height in meter ) . (take g =10 m/s^2)

Answers

Answered by Fatimakincsem
20

The height of ball 1 is 15 m, of ball 2 is 20 m and of ball 3 is 20 m respectively.

Explanation:

Time taken by same ball to return to the hands of juggler =2u / g

                                                                                               = 2×20 / 10 = 4s.

  • Therefore he is throwing the balls after each 1s.
  • Let at some instant he is throwing ball number 4.
  • Before 1s of it he throws ball 3.

So height of ball 3

h3 = 20 × 1 − 1 / 2 x 10(1)^2 = 15 m

Before 2s, he throws ball 2. So height of ball 2 is

h2 = 20 × 2 − 1 / 2 x 10(2)^2 = 20 m

Before 3s, he throws ball 1. So height of ball 1 :

h1 = 20 × 3 − 1 / 2 x 10(2)^2 = 20 m

Hence the height of ball 1 is 15 m, of ball 2 is 20 m and of ball 3 is 20 m respectively.

Answered by Avikingj
15

Here is ur ans

Answer

The height of ball 1 is 15 m, of ball 2 is 20 m and of ball 3 is 20 m respectively.

Explanation:

Time taken by same ball to return to the hands of juggler =2u / g

= 2×20 / 10 = 4s.

Therefore he is throwing the balls after each 1s.

Let at some instant he is throwing ball number 4.

Before 1s of it he throws ball 3.

So height of ball 3

h3 = 20 × 1 − 1 / 2 x 10(1)^2 = 15 m

Before 2s, he throws ball 2. So height of ball 2 is

h2 = 20 × 2 − 1 / 2 x 10(2)^2 = 20 m

Before 3s, he throws ball 1. So height of ball 1 :

h1 = 20 × 3 − 1 / 2 x 10(2)^2 = 20 m

Hence the height of ball 1 is 15 m, of ball 2 is 20 m and of ball 3 is 20 m respectively.

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