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:

For vertical projection, maximum height is \(h=\frac{u^2}{2g}\) and time to highest point is \(t=\frac{u}{g}\).

Answer 1

The Correct Option is A

Concept:
For a vertically projected body, the time to reach maximum height is: \[ t=\frac{u}{g} \] and maximum height is: \[ h=\frac{u^2}{2g} \]

Step 1: The juggler throws \(n\) balls per second.

Step 2: Therefore, time interval between two throws is: \[ t=\frac{1}{n} \]

Step 3: He throws the next ball when the previous ball is at the highest point. Hence, time to reach the highest point is: \[ t=\frac{1}{n} \]

Step 4: Since: \[ t=\frac{u}{g} \] \[ u=gt=\frac{g}{n} \]

Step 5: Maximum height: \[ h=\frac{u^2}{2g} \] \[ h=\frac{\left(\frac{g}{n}\right)^2}{2g} \] \[ h=\frac{g^2}{n^2}\cdot\frac{1}{2g} \] \[ h=\frac{g}{2n^2} \] \[ \boxed{\frac{g}{2n^2}} \]