Question

A student is throwing balls vertically upwards such that he throws the $2^{\text{nd}}$ ball when the $1^{\text{st}}$ ball reaches maximum height. If he throws balls at an interval of 3 second, the maximum height of the balls is ($g = 10\ \text{ms}^{-2}$)

• A. 45 m
• B. 35 m
• C. 25 m
• D. 30 m

Hint:

For any object falling freely from rest, the distance covered in the first few seconds follows the sequence $5t^2$. Plucking in $t=3$ directly yields $5 \times 9 = 45\ \text{m}$ instantly.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A person throws balls vertically upward at a steady timeline spacing of $\Delta t = 3\ \text{s}$. The second ball is released at the exact instant the first ball hits its peak altitude. We need to determine this maximum height.

Step 2: Key Formula or Approach:
1. The time of ascent ($t_{\text{ascent}}$) to reach maximum height under uniform gravitational acceleration is equal to the interval between throws.
2. From kinematic equations, the distance traveled by an object falling freely from rest from that peak height over the same time interval is:
$$H = \frac{1}{2}gt^2$$

Step 3: Detailed Explanation:
Since the second ball is thrown exactly when the first ball reaches its highest point, the time taken for any ball to reach its maximum height is exactly equal to the throwing interval:
$$t_{\text{ascent}} = 3\ \text{s}$$ By the symmetry of vertical motion under gravity, the time taken to climb to the top equals the time taken to fall back down from that peak to the launch point. Let's track the downward trip as a free fall from rest ($u=0$) lasting for $t = 3\ \text{s}$:
$$H = \frac{1}{2}gt^2$$ Substitute $g = 10\ \text{ms}^{-2}$ and $t = 3\ \text{s}$:
$$H = \frac{1}{2} \times 10 \times (3)^2$$ $$H = 5 \times 9 = 45\ \text{m}$$ This calculation establishes the maximum height reached by the balls.

Step 4: Final Answer:
The maximum height of the balls is 45 m, which matches option (A).