A clown in circus juggles with n balls using only one hand. He throws each ball vertically upwards with the same speed at equal time intervals. Denote acceleration of free fall by g.

(a) Find expressions for the speed of projection and height of the i th ball above his hand when he throws the n th ball.
Let he uses n = 4 balls and when he throws the fourth ball, the distance between the second and third ball is d = 50 cm.
(b) Where is the first ball, when the juggler throws the fourth ball?
(c) What is maximum height attained by each ball above the hands of the juggler?

To tackle this problem, we need to break it down into manageable parts. The juggler throws each ball vertically upwards with the same speed at equal time intervals. Let's denote the speed of projection as $v$ and the time interval between throws as $t$. The acceleration due to gravity is represented by $g$. We will analyze the motion of the balls step by step.

Speed of Projection and Height of the i-th Ball

When the juggler throws the n-th ball, the i-th ball will have been in the air for a certain amount of time. The time each ball spends in the air can be calculated based on the time interval $t$ and the number of balls thrown before it. For the n-th ball, the i-th ball will have been in the air for $(n-i)\cdot t$ seconds.

The speed of projection $v$ can be expressed in terms of the distance $d$ between the second and third balls when the fourth ball is thrown. Since the distance between the balls is given as $d=50$ cm, we can use the kinematic equations to find $v$.

Finding the Speed of Projection

The vertical distance $h$ traveled by a ball after time $t$ is given by the equation:

• $h=vt-\frac{1}{2}g{t}^2$

For the second and third balls, the time intervals are $t$ and $2t$ respectively. The distance between them can be expressed as:

• $h_3-h_2=(v(2t)-\frac{1}{2}g(2t{)}^2)-(v(t)-\frac{1}{2}g{t}^2)$

Substituting the distances:

• $d=v(2t)-\frac{1}{2}g(4t^2)-(v(t)-\frac{1}{2}g{t}^2)$

After simplifying, we find:

• $d=vt-\frac{3}{2}g{t}^2$

Given $d=50$ cm (or 0.5 m), we can rearrange this to find $v$:

• $v=\frac{d+\frac{3}{2}g{t}^2}{t}$

Height of the i-th Ball

Now, to find the height of the i-th ball when the n-th ball is thrown, we use:

• $h_i=v(n-i)t-\frac{1}{2}g(n-i{)}^2{t}^2$

Position of the First Ball

When the juggler throws the fourth ball, the first ball has been in the air for $3t$. The height of the first ball can be calculated using the same kinematic equation:

• $h_1=v(3t)-\frac{1}{2}g(3t{)}^2$

Maximum Height of Each Ball

The maximum height attained by each ball occurs when its velocity becomes zero. The time to reach maximum height for each ball can be calculated as:

• $t_{max}=\frac{v}{g}$

The maximum height $H$ for each ball can then be calculated using:

• $H=v{t}_{max}-\frac{1}{2}g{t}_{max}^2$

Substituting $t_{max}$ into the equation gives:

• $H=\frac{v^2}{2g}$

In summary, we have derived the expressions for the speed of projection, the height of the i-th ball, the position of the first ball when the fourth ball is thrown, and the maximum height attained by each ball. This systematic approach allows us to understand the dynamics of juggling in a clear and logical manner.