The juggler and his balls

let the time taken by each ball to reach max height be $t$ . Since ball is at rest on the top...final velocity $v=0$ .

Thus $\begin{aligned} v&=u-gt\\ \implies 0&=u-gt\\ \implies t&=\dfrac { u }{ g } \\ \end{aligned}$

Now, $t$ is the time taken by $1$ ball. So time taken by $n$ balls = $\dfrac{nu}{g}$

We know that $n$ balls take $1 sec$ . Thus $\dfrac{nu}{g}=1\\ \implies u=\dfrac{g}{n}$

Now applying $v^2 = u^2 - 2gs$ , we get

$\begin{aligned} 0&=\left( \dfrac { { g }^{ 2 } }{ { n }^{ 2 } } \right) -2gs\\ \implies 2gs&=\left( \dfrac { { g }^{ 2 } }{ { n }^{ 2 } } \right) \\ \implies 2s&=\dfrac { g }{ { n }^{ 2 } } \\ \implies s&=\boxed{\dfrac { g }{ 2{ n }^{ 2 } }} \end{aligned}$

A ball is thrown every 0.5 seconds. So the previous ball (indeed every ball) takes 0.5 seconds to reach its peak. Assuming we can neglect air resistance, the ball takes the same time to rise to its peak as it does to fall from its peak back to the juggler's hand.

At its peak the ball is momentarily stationary and so it is easy to calculate the distance it falls in 0.5s as

$s=ut+\dfrac{1}{2}at^2=0+\dfrac{1}{2}\times 10\times 0.5^2=1.25$

Either using a little optimization calculus, or the formula Vf= V1 + at, it is easy to see that the time is v/g. "n" that is given in the problem is the frequency of balls per second. Frequency is the inverse of period. If n=2, then t=1/2. Substituting this value for t, in the previous formula, t=v/g, (g=10 m/s^2) velocity clearly equals 5m/s. Distance is -1/2gt^2 +vt.

-5(1/2)^2 + 5(1/2) =1.25

Every second 2 balls going up means a ball takes 1 sec for going up & coming down. It means free falling ball takes 1/2 sec . : the distance travelled in 1/2 sec is s = 1/2 *gt^2 =10/(2 X4 )=1.25

There's a formula for the condition given the question-:

$H = \dfrac{g}{2n^{2}}$

Putting values given in the question,we get our answer as $1.25$ .

Two balls thrown every second means at the start of the second a ball is thrown, and when it reaches the highest point another is thrown and thats the end of the second so the time taken to reach the highest point must be 1 second. By v=u+at.. U =10m/s.. h=u^2/2g... H=5m... Please tell me whats wrong in this.

when 1 ball is thrown up at the same time the another ball is released from rest.i.e;the maximum height reached by the ball thrown from down is equal to the time taken by the upper ball to come down.i.e; for ball thrown up final velocity=0,using v=u-gt we get u=gt for n balls nu=gt here time =1sec,g=10,n=2 so u=5 now,max. height reached: using,v2-u2=2gh we get h=u2/2g=25/20=1.25

here u2=u square&v2=v square

use H=g/2n^2