Question

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

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

Step-by-step solution

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.

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 \(t_B\): \(t_B = \sqrt{\frac{2h}{g}}\)

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) \(\sqrt{(2\mathrm{~h} / \mathrm{g})}\).

Key concepts

Equations of Motion

Equations of motion are vital when analyzing objects in projectile motion, like the balls in our exercise. These equations allow us to predict how objects move under the influence of forces. Essential equations include:

• First Equation: \( v = u + at \) - \( v \) is the final velocity. - \( u \) is the initial velocity. - \( a \) is the acceleration. - \( t \) is time.
• Second Equation: \( s = ut + \frac{1}{2}at^2 \) - \( s \) is the displacement.
• Third Equation: \( v^2 = u^2 + 2as \)

In our scenario, these equations allow us to determine the time each ball takes to travel through their respective motions. Specifically, the equation \( v = u - gt \) assumes the upward or downward movement along the y-axis, with gravity acting downwards. This helps us identify the time when ball A reaches its maximum height, as we adjusted the equation to find \( t_A \) by setting final velocity (\( v \)) as zero.

Acceleration Due to Gravity

Acceleration due to gravity, denoted by \( g \), is one of the most important constants in physics. This value represents the acceleration that any object in free fall experiences due to the gravitational pull of Earth. Its average value is approximately \( 9.81 \, \text{m/s}^2 \) on the surface of the Earth.
In projectile motion problems, gravity acts as an acceleration component that affects the velocity and trajectory of moving objects.

For ball A, thrown upwards, gravity works to slow it down until it reaches its peak and begins falling back. Meanwhile, ball B, dropped from rest, accelerates downwards due to gravity, which increases its velocity as it approaches the ground. Using the equation \( h = \frac{1}{2}gt^2 \), we find the time needed for ball B to fall, reflecting how gravity alone influences the fall, with no initial velocity involved.

Kinematics

Kinematics is the study of motion without considering the forces that cause motion. It helps us understand projectile motion by focusing on variables such as displacement, velocity, acceleration, and time. In problems like the one we're solving, kinematic equations are used to calculate these variables.

In the context of our exercise, kinematics allows us to analyze and predict the behavior of both balls. We focus on:

• Displacement—the height of the tower, an initial condition influencing both particles.
• Velocity—the changes in speed as it interacts with gravity, calculated to understand how each move through time.
• Time of Flight—how long each ball is in motion. Notably, ball A reaching its maximum height simultaneously when ball B hits the ground.

Through kinematics, we can use equations like \( t = \sqrt{\frac{2h}{g}} \) to find the time intervals in which the balls complete their motions, respecting their initial conditions and the uniform influence of gravity.