91Ó°ÊÓ

Gravitational fields are regions of space where objects with mass experience a force due to gravity. This force pulls objects towards the center of mass exerting the gravitational attraction. On Earth, we commonly express this in terms of acceleration, generally denoted as gravitational acceleration, symbolized by \( g \). Here, \( g = 10 \, \mathrm{m/s^2} \), as used in our problem. A notable aspect of motion in a gravitational field is that all objects, regardless of mass, will experience the same rate of acceleration.

• This constant gravitational force is what causes objects to fall when dropped.
• While air resistance may alter the real-world case, it’s often neglected in calculations for simplicity.
• That means when an object is dropped from rest, its motion relies solely on the gravitational pull.

Understanding how objects move under the influence of gravity is fundamental in physics. It allows us to predict their behavior, particularly in free-fall situations where other forces are minimal or negligible.

Free-fall motion occurs when the only force acting on an object is gravity. In our exercise, a ball is dropped from a height, entering a free-fall state. This means it starts with no initial velocity and accelerates downwards at \( g = 10 \, \mathrm{m/s^2} \). Free-fall motion is described by kinematic equations, such as \( h = \frac{1}{2} g t^2 \).

• In every second of free-fall, the speed of the ball increases by \( 10 \, \mathrm{m/s} \).
• The time it takes for the ball to hit the ground from a height \( h \) can be calculated using the time equation derived from the kinematic formula.
• To find out how long it takes to drop 45 meters, we solve \( 45 = \frac{1}{2} \times 10 \times t^2 \) resulting in \( t = 3 \, \mathrm{s} \).

This calculated time only considers the downward motion to the ground without an impact or bounce consideration, which is an essential aspect in problems involving free-fall and rebounds.

Rebound height involves understanding the behavior of an object after it strikes a surface and bounces back. The coefficient of restitution \( e \) plays a vital role here. It’s a measure of how much energy of motion (kinetic energy) is conserved during a collision. For the ball in our problem, \( e = \frac{2}{3} \).

• This value means the ball retains \( \left( \frac{2}{3} \right)^2 \) of its motion energy when it rebounds.
• Thus, to calculate the height of the rebound, we use the formula \( h' = e^2 \times h \). Here, \( h = 45 \, \mathrm{m} \), making \( h' = 20 \, \mathrm{m} \).
• After calculating the initial rebound height, we can determine the time taken for further motion. Using the formula \( h' = \frac{1}{2} g t'^2 \), we get \( t' = 2 \, \mathrm{s} \).

In our exercise, understanding rebound height and time contributes to assessing the distance traveled during specific seconds of motion, providing insights into periods like the fourth second, where the net distance of \( 5 \, \mathrm{m} \) was established.