Question

A juggler throws balls into air. He throws one ball whenever the previous one is at its highest point. If he throws 4 balls in one second, the height from which each ball returns is nearly [acceleration due to gravity is \(9.8\text{ m/s}^{2}\)]

• A. \(30\text{ cm}\)
• B. \(40\text{ cm}\)
• C. \(50\text{ cm}\)
• D. \(60\text{ cm}\)

Hint:

The time between throws equals the time a ball takes to reach its highest point. Use the free fall formula for that time.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
A new ball is thrown when the previous ball is at its highest point. So the time gap between throws equals the time of ascent.

Step 2: Find the Time
4 balls are thrown in 1 s, so the gap is \(t=\dfrac{1}{4}=0.25\text{ s}\).

Step 3: Calculation
At the top the speed is zero, so the fall from the top takes the same \(0.25\text{ s}\).
\(h=\dfrac{1}{2}gt^{2}=\dfrac{1}{2}(9.8)(0.25)^{2}\approx0.306\text{ m}\)

Step 4: Compare the options
\(0.306\text{ m}\approx30\text{ cm}\), which matches the first option.

Final Answer:
Each ball rises to nearly 30 cm. \[ \boxed{\text{(A) }30\text{ cm}} \]