This juggler problem drining me crazy

[Alpharup]

Homework Statement

A juggler throws balls up with wqual speed. He releases a ball when the previous pne has reached maximum height. the number of balls moving in the upward direction at one instant is
[a]1 2 [c]3 [d]4

Homework Equations

Let the velocity of the balls thrown be 'u'. Let the maximum height that the balls reach be 'h'. Let the time difference between release of two balls be 't'. let the time to reach the maximum height be 'T'

The Attempt at a Solution

h=uT-[1/2]gT^2
[h/2]=ut-[1/2]gt^2

 

[omoplata]

h=uT-[1/2]gT^2
[h/2]=ut-[1/2]gt^2

I agree with the first equation. But how did you get the second one? The first ball has reached a height h before the second ball is released, not h/2.

 

[Vanadium 50]

Try it without equations. Draw diagrams of what's happening.

 

[sankalpmittal]

Homework Statement

A juggler throws balls up with equal speed. He releases a ball when the previous one has reached maximum height. the number of balls moving in the upward direction at one instant is
[a]1 2 [c]3 [d]4

Homework Equations

Let the velocity of the balls thrown be 'u'. Let the maximum height that the balls reach be 'h'. Let the time difference between release of two balls be 't'. let the time to reach the maximum height be 'T'

The Attempt at a Solution

h=uT-[1/2]gT^2
[h/2]=ut-[1/2]gt^2

Homework Statement

Homework Equations

The Attempt at a Solution

Ok , let's move step by step.

Let he throw a ball at u velocity as you mentioned. Then what is the maximum height attained by that ball ? Use this equation of motion :

h= 1/2t(v+u)

Now what is the time interval when that ball has reached its height h ? t right ? Then what will be the time interval between that ball attaining h height and release of another ball ?

[Your second equation is absurd I think. How did you get it ?]

 

[rwisz]

t^2-2uT+2h/g=0
t= [2uT+sqrt(4u^2T^2-8gh)]/2
t=uT+sqrt(u^2T^2-2gh)
t=uT+sqrt(0)
t=uT

Since the time difference between release of two balls is equal to the time taken for a ball to reach maximum height, the number of balls moving in the upward direction at one instant will always be 1. Therefore, the correct answer is [a]1.

I would also like to mention that this problem can be solved using the equations of motion and the concept of projectile motion. The number of balls moving in the upward direction at one instant can also be determined by analyzing the velocity and acceleration of the balls at different points in time. I would suggest practicing more problems on projectile motion to gain a better understanding of this concept.