Question

A juggler throws balls into the air. He throws one whenever the previous one is at its highest point. How high do the halls rise if he throws $n$ balls each second ?

Answer 1

Hint:When a ball is thrown upwards then there is some initial velocity $u$ and a final velocity $v$. The final velocity becomes zero at the highest point. And the force which works on the ball in this situation is the gravitational force, hence the acceleration a becomes -g(acceleration due to gravity is in opposite direction of motion when ball goes from ground to top).

Complete step by step answer:
Given that the juggler throws $n$ balls in one second. It implies that one ball takes \[\dfrac{1}{n}\] seconds to reach at the highest point. Three equations of motion are:
First equation: \[v~=~u~+at\]
Second equation: \[s=ut+\dfrac{1}{2}a{{t}^{2}}\]
Third equation: \[2as={{v}^{2}}-{{u}^{2}}\]
Let’s this time be t seconds, i.e. \[t=\dfrac{1}{n}\].

When a ball is thrown upwards then there is some initial velocity u and a final velocity $v$.The final velocity becomes zero at the highest point. And the force which works on the ball in this situation is the gravitational force, hence the acceleration a becomes -g(acceleration due to gravity is in opposite direction of motion when ball goes from ground to top).

Now putting \[v=0\] and \[a=-g\]in the first equation of motion, \[v=u+at\], we get: \[u=\dfrac{g}{n}\]. It means the initial velocity with which each ball is thrown upwards is \[\dfrac{g}{n}\]. Now let’s consider the motion to the highest point; initial velocity \[u=\dfrac{g}{n}\], final velocity \[v=0\], total distance (s)= maximum height that a ball gains (h). Applying third equation of motion and putting the above quantities, we observe that:
\[2as={{v}^{2}}-{{u}^{2}}\] changes to \[2gh={{0}^{2}}-\dfrac{{{g}^{2}}}{{{n}^{2}}}\].
\[\therefore h=\dfrac{g}{2{{n}^{2}}}\]

Therefore the maximum height taken by each ball when a juggler throws n balls in one second and also each ball is thrown whenever the previous one is at its highest point is \[\dfrac{g}{2{{n}^{2}}}\].

Note:According to the question $n$ balls are thrown each second, which means $n$ balls are thrown in one second. Then by unitary method we can say that one ball is thrown in \[\dfrac{1}{n}\] seconds. You can also directly calculate, without using the unitary method.