Understanding the Downward Speed of a Basketball: A Physics Exploration

Ever found yourself pondering what happens when you drop a basketball? More specifically, how fast that basketball is descending at a specific height? Let's break it down using the principles of energy conservation—a crucial concept for anyone gearing up for the UCF PSC1121 Physical Science Final Exam. When you let go of that basketball, it starts its journey from rest, and as it falls, it picks up speed. So, what’s its downward speed when reaching a height of 2.50 meters? You might be surprised by how elegantly physics handles this!

To determine the downward speed of our basketball at that 2.50-meter mark, we tap into the idea of gravitational potential energy transforming into kinetic energy. It’s like watching a thrilling movie where every moment of suspense escalates until the grand reveal. The exciting twist here? The reveal is all about energy exchanges!

Now, let’s lay out the groundwork: the gravitational potential energy (PE) of the basketball is calculated using the formula: [ PE = mgh ] In this equation, ( m ) stands for the mass of the basketball, ( g ) is the acceleration due to gravity (which you can always count on to be approximately ( 9.81 , \text{m/s}^2 )), and ( h ) is the height—2.50 m in this case.

So why is this important? Well, as the basketball falls, it trades that potential energy for kinetic energy (KE), which you can calculate using: [ KE = \frac{1}{2} mv^2 ] That moment when the basketball hits 2.50 m is where the magic happens. It's converting potential energy into kinetic energy, just like a swimmer diving off a platform into a pool below—it’s the same basic principle!

Here’s the crux: at that height of 2.50 m, we equate the potential energy with the kinetic energy: [ mgh = \frac{1}{2} mv^2 ] Since the mass ( m ) appears on both sides of the equation, it gracefully cancels itself out, leaving us with a clean path to the solution. Let's tackle the math without losing our momentum. Rearranging the equation gives us: [ v^2 = 2gh ] Now substitute the numbers: [ g \approx 9.81 , \text{m/s}^2 , \text{and} , h = 2.50 , \text{m} ] Doing the math results in: [ v^2 = 2(9.81)(2.50) \implies v^2 \approx 49.05 ] To find ( v ), we take the square root: [ v \approx 7.00 , \text{m/s} ] And there you have it! The speed of the basketball at a height of 2.50 m is approximately 7.00 m/s—not just another number, but a reflection of the power of physics.

Getting the hang of concepts like potential and kinetic energy not only shows you how intriguing forces like gravity can be but also helps you solve problems more effectively in your final exam. So next time you shoot hoops, remember the physics at play! Learning doesn’t stop here—how about exploring other concepts related to energy? The world is buzzing with scientific wonders waiting for your curious mind!