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Chapter 2: Problem 199

Ball A is thrown in upward from the top of a tower of height h. At the same time ball B starts to fall from that point. When A comes to the top of the tower, B reaches the ground. Find the time to reach maximum height for \(\mathrm{A}\). (A) \(\sqrt{(\mathrm{h} / \mathrm{g})}\) (B) \(\sqrt{(2 \mathrm{~h} / \mathrm{g})}\) (C) \(\sqrt{(h / 2 g)}\) (D) \(\sqrt{(4 \mathrm{~h} / \mathrm{g})}\)

Short Answer

Expert verified
(B) \(\sqrt{(2\mathrm{~h} / \mathrm{g})}\)

Step by step solution

01

Identify relevant equations

For ball A moving upwards, we will use the following equation of motion: \(v = u - gt\) Here, \(v\) represents final velocity, \(u\) represents initial velocity, \(g\) represents acceleration due to gravity, and \(t\) represents time. Similarly, for ball B moving downwards, we will use the following equation of motion: \(h = \frac{1}{2}gt^2\) Now, we know that at maximum height, the velocity of ball A will be zero. Therefore, we can rewrite the first equation for ball A as: \(0 = u - gt_A\) To find the time taken for ball A to reach its maximum height (\(t_A\)), we will first need to find the time taken for ball B to reach the ground (\(t_B\)) and then compare both times.
02

Find the time taken for ball B to reach the ground

Use the equation of motion for ball B: \(h = \frac{1}{2}gt_B^2\) Solve for \(t_B\): \(t_B = \sqrt{\frac{2h}{g}}\)
03

Find the time taken for ball A to reach its maximum height using the fact that both balls reach their extreme points at the same time

Since ball A comes to the top of the tower when ball B reaches the ground, we have: \(t_A = t_B\) Thus, \(t_A = \sqrt{\frac{2h}{g}}\) The correct answer is (B) \(\sqrt{(2\mathrm{~h} / \mathrm{g})}\).

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