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:

If \(n\) balls are thrown per second, time interval is \(\frac{1}{n}\). This is the time to reach the highest point.

Answer 1

The Correct Option is A

The juggler throws \(n\) balls each second. Therefore, the time interval between two successive throws is: \[ t=\frac{1}{n}. \] He throws one ball whenever the previous ball reaches its highest point. So, the time taken by a ball to reach the highest point is: \[ t=\frac{1}{n}. \] At the highest point, final velocity becomes zero: \[ v=0. \] Let the initial velocity of projection be \(u\). Using: \[ v=u-gt. \] At the highest point: \[ 0=u-gt. \] So, \[ u=gt. \] Substitute: \[ t=\frac{1}{n}. \] \[ u=\frac{g}{n}. \] Now maximum height is: \[ h=\frac{u^2}{2g}. \] Substitute: \[ u=\frac{g}{n}. \] \[ h=\frac{\left(\frac{g}{n}\right)^2}{2g}. \] \[ h=\frac{\frac{g^2}{n^2}}{2g}. \] \[ h=\frac{g}{2n^2}. \] Hence, the balls rise to height: \[ \frac{g}{2n^2}. \]