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

Hint:

We use the equations of rectilinear motion to calculate the total time of flight of each ball to find the positions of the balls with respect to the ball in the hand of the juggler. 

Answer 1

The Correct Option is B

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.

Answer 2

We use the equations of rectilinear motion to calculate the total time of flight of each ball to find the positions of the balls with respect to the ball in the hand of the juggler. 

The equation for rectilinear motion under gravitational field h = ut + ½ gt², 

• where u is the initial velocity of the object,
• t is the total time of fall or fly, and
• g is the acceleration of the particle

Complete Step-By-Step Answer:

The juggler juggles four balls together and we know that the initial velocity of the balls when it leaves the juggler’s hand is 20 ms^-1. 

Then, the total time of flight of each ball (t) can be found using the formula, h = ut + ½ gt², where h is 0 as the ball goes up and comes back to the hand.

Putting the value of u = 20 ms^-1 and g = 10ms^-2 we get, \(0=20t−\frac{1}{2}gt^2\)

Solving this equation we get,

\(\frac{1}{2}gt^2=20t\)

\(t(gt−40)=0\)

\(gt=40\)

Since t cannot be zero as the ball goes up and come back.

Therefore, \(t=\frac{40}{g}=\frac{40}{10}=4s\), given \(g=10ms^{−2}\)

So, each ball goes up and comes back in 4s. Hence, after each second one of the balls comes back and the other three are in the air. So, the separation between each ball is 1s.

Let's say that at some point, the 4th ball comes back, the other three must be in the air then.

• So, the third ball already has left the hand 1s before the fourth ball.
• Its height will be the height reached by a particle in 1s if it were thrown up.
• In a similar manner, the second ball left the hand 2s before, and the first ball left 3s ago.

Let’s find the height of each ball: 

Height of the fourth ball: \(t=1s\) and \(u=20ms^{−1}\), \(g=10ms^{−2}\).

Height of the third ball: \(h=20×1−\frac{1}{2}10×1^2\) \(h=20−5=15m\) from the hand of the juggler.

Height of the second ball: : \(t=2s\) and \(u=20ms^{−1}\), \(g=10ms^{−2}\),  \(h=20×2−\frac{1}{2}10×2^{2}\)  Or, \(h=40−20=20m\)

Height of the first ball:: \(t=3s\) and \(u=20ms^{−1}\), \(g=10ms^{−2}\)

Therefore, \(h=20×3−\frac{1}{2}10×3^2\)  Or, \(h=60−45=15m\)

Therefore the balls will be at a separation of 15,20,15 metres from the hand of the juggler.

Hence, option (B) is correct.

Note: If the juggler juggles the balls with his two hands but separately, then the separation between the balls would be 4s.

To juggle any number of balls the separation between each ball must be symmetrical or else the balls cannot be juggled.