Question

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

• A. \(\frac{g}{2n^2}\)
• B. \(\frac{g}{n}\)
• C. \(\frac{g}{2n}\)
• D. \(\frac{n^2}{g}\)

Hint:

This problem connects rate (\(n\) balls per second) to time (\(t = 1/n\)). Once the time to reach the peak is known, the kinematics of the projectile can be fully determined. Breaking down the problem statement into physical quantities is the first crucial step.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A juggler throws \(n\) balls per second. The time interval between throws is the time it takes for a ball to reach its maximum height. We need to find this maximum height.

Step 2: Key Formula or Approach:
1. Determine the time of flight to the highest point.
2. Use kinematic equations to relate this time to the initial velocity (\(v=u+at\)).
3. Use another kinematic equation to relate the initial velocity to the maximum height (\(v^2=u^2+2as\)).

Step 3: Detailed Explanation:
If the juggler throws \(n\) balls each second, the time interval between two consecutive throws is \(\Delta t = \frac{1}{n}\) seconds.
The problem states this is the time for a ball to reach its highest point. Let's call this time \(t_{up}\).
\[ t_{up} = \frac{1}{n} \]
At the maximum height, the final vertical velocity \(v\) is 0. Using \(v = u + at\) with \(a = -g\) (upwards as positive):
\[ 0 = u - g \cdot t_{up} \]
\[ u = g \cdot t_{up} = g \cdot \frac{1}{n} = \frac{g}{n} \]
This is the initial velocity with which each ball is thrown.
Now, to find the maximum height \(H\), we use the equation \(v^2 = u^2 + 2as\):
\[ 0^2 = u^2 + 2(-g)H \]
\[ u^2 = 2gH \]
\[ H = \frac{u^2}{2g} \]
Substitute the expression for \(u\) we found:
\[ H = \frac{(g/n)^2}{2g} = \frac{g^2/n^2}{2g} = \frac{g}{2n^2} \]

Step 4: Final Answer:
The balls rise to a height of \(\frac{g}{2n^2}\).