Question

A juggler throws balls vertically upwards with same initial velocity in air. When the first ball reaches its highest position, he throws the next ball. Assuming the juggler throws n balls per second, the maximum height the balls can reach is

• A. g/2n
• B. g/n
• C. 2gn
• D. g/2n2

Answer 1

The Correct Option is D

To solve this problem, we will analyze the motion of the balls thrown by the juggler.

Let's denote: 

• Initial velocity with which each ball is thrown as \(u\).
• Acceleration due to gravity as \(g\) (acting downwards).
• Time taken for the ball to reach the maximum height as \(t\).

According to the problem, the juggler throws the next ball when the first ball reaches its highest position, which implies that \(n\) balls are thrown per second.

For a ball thrown upwards, at the highest point, the velocity becomes zero. The time to reach this point is given by the first equation of motion:

\(v = u - gt\)

At the highest point, \(v = 0\), so:

\(0 = u - gt\)

\(u = gt\)

The ball should reach the maximum height by the time another ball is thrown, so:

\(t = \frac{1}{n}\), as \(n\) balls are thrown per second.

Substitute \(t = \frac{1}{n}\) into \(u = gt\):

\(u = g\left(\frac{1}{n}\right) = \frac{g}{n}\)

Now calculate the maximum height \(H\) using the formula:

\(H = \frac{u^2}{2g}\)

Substitute \(u = \frac{g}{n}\) into the equation:

\(H = \frac{\left(\frac{g}{n}\right)^2}{2g}\)

Simplify:

\(H = \frac{g^2}{2gn^2}\)

\(H = \frac{g}{2n^2}\)

Therefore, the maximum height the balls can reach is \(\frac{g}{2n^2}\).

Hence, the correct option is \(\frac{g}{2n^2}\).

Answer 2

\(t=\frac{u}{g}=\frac{1}{n}\)
Then,
u=\(\frac{g}{n}\)
H_max=\(\frac{u^2}{2g}=\frac{g}{2n^2}\)
So, the correct option is (D): g/2n²