I'm sitting on my warm, snuggly couch watching the Cleveland Indians and the Chicago Cubs in Game 2 of the World Series. The players, coaches, and fans are shivering on my television screen. Today, we are going to discuss the physics of home runs; unfortunately, the cold weather makes it unlikely anyone will hit a dinger in tonight's game. This phenomenon is rooted in fluid dynamics. Cold air has a higher density which results in increased drag on the ball. However, that discussion is outside the scope of this article. We are going to investigate how launch angle impacts whether you flyout or jog around the bases. Recall that the trajectory of a falling object is parabolic; meaning, it has an x and a ycomponent. Knowing that distance travelled by a falling object is related to the initial velocity (v), we can mathematically express each component of the velocity utilizing concepts from trigonometry: We get the following expressions after plugging the expressions above into the formulas for a zero launch angle: Now we can really start digging into our problem. Let's make the following assumptions:
You may notice we don't have a time component, which is a problem since we have 3 unknowns and two equations. However, we can overcome this by reframing our question: How far has the ball travelled when it reaches a height of 3 m? We ask this question because we want to find the ideal launch angle to hit a home run. So the mechanics look like this: Therefore, under the assumed conditions a launch angle of 45 degrees would result in a home run! Let's look at a range of angles to find the "sweet spot:" It appears when the ball is traveling at 35 m/s the ideal launch angle is between 45 and 50 degrees. The next question is: How would this change with varying initial velocities? As expected, you have greater flexibility of launch angle with increasing velocity. So if you want to crush the ball, you better start pumping iron!
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This week we find ourselves in the middle of the Major League Baseball (MLB) Post Season. In the flurry of excitement we hear commentators spouting off different stats. Since everyone "digs the long ball" you better believe you are hearing exit velocity and pitch speed stats. The exit velocity is the speed of the ball coming off the bat and  quite obviously  the pitch speed is the speed of the incoming throw from the pitcher. These stats are important elements of predicting the characteristic of the hit ball, particularly the ball's distance travelled. However, there has been much debate on the importance of pitch speed on the ultimate exit velocity. Some argue the faster the ball comes in the faster the ball will go out. However, this is not the case. To understand the impact of pitch velocity we need to look at the momentum of the ball. In this system we consider the forces acting on the ball. Consider the setup below: The impulse (I) is the average force (Fav) delivered to the ball over time (delta t). This is also considered the change in momentum (delta p). The change in momentum can also be expressed as the initial momentum subtracted from the resulting final momentum. When we combine these two equations with the formula for momentum (mass multiplied by velocity) we see a relationship between the force delivered by the batter (Fav) and the speed of the pitch (vpitch). Since we are primarily interested in how these pieces interact to influence the exit velocity, let's isolate the exit velocity: You may be thinking to yourself, "Whoa, the pitch velocity is added to the force applied by the batter. So faster pitches should lead to faster exit velocities." However, I would remind you to remember our coordinate system. If our pitch velocity is 90mph, we would write vpitch =  90mph since it is traveling in the negative x direction. Therefore, pitch velocity decreases the ultimate exit velocity. To give you an example, let's assume two pitches are thrown and the batter applies the same force to each pitch. How would the exit velocity differ? (We will assume the Fav term will equal 200mph for the ease of the example.) This phenomenon can be observed in the MLB stats. Visit Statcast to see that some of this season's hardest hit balls were on relatively slow pitches. Another theory MLB player's have is that since the ball is traveling slower the batter has a better chance at hitting the ball square. There is some truth to this in the likelihood of turning a pitch into a "dinger." However, the physics shows us that if both balls are hit in the same manner that the ball hit off the slower pitch will go faster.
Enjoy the rest of the Post Season with this tidbit of information. Ask your friends watching the game what they think would result in a harder hit ball, then impress them with your insight. Go Cubs! :P It happens to all of us. We reach into the fridge and pull out the ketchup bottle only to that find the remainder of ketchup is clinging for dear life at the bottom of the bottle. There are two well known practices of getting the remainder of the ketchup out of the bottle. Either you place the bottle on its head and wait for gravity to do its job, or you shake the ketchup down. But, I'm here to tell you there is a third and more efficient way to get that last bit of ketchup out of the bottle. Centripetal force is plainly defined at the force exerted on an object towards the center when moving in a circular motion. Centrifugal force is the opposing force exerted on an object to move it away from the center. Both of these forces are required to understand what is happening in our ketchup bottle.
This causes an unbalanced force to act on the ketchup. Remember what Newton's First Law states? "An object at rest tends to stay at rest and an object in motion tends to stay in motion in the same direction and speed unless acted upon by an unbalanced force." Therefore, the ketchup accelerates away from the center of circular motion until it hits the lid, which subsequently exerts the centripetal force towards the center of the circular motion to stop the flow of ketchup.
Give this technique a "whirl" the next time you need to get the remaining bit of ketchup out of the bottle and impress your friends and family with your knowledge of Physics! Have you ever wondered what causes your favorite instrument to create sound? The answer is rooted in the physics of waves. There are many different types of waves: transverse, longitudinal, and standing. Transverse waves move perpendicular to the direction of the wave, longitudinal waves move parallel to the direction of the wave, and standing waves appear stationary in space. Basic elements that define a wave are its wavelength, period/frequency, and speed. A wavelength (w) is the distance over which a wave repeats itself. The period (T) is the time required for a single wavelength to pass a defined point. The frequency (f) is inversely related to the period. Finally, speed (v) is the same as if we were referring to the speed of a car: the amount of distance covered over time. Instruments produce sounds via standing waves. Standing waves are composed of nodes (the point at which the wave appears to have zero displacement) and antinodes (the point at which the wave appears to have maximum displacement). The frequency  otherwise known as a harmonic for standing waves  is determined by the fundamental (first) harmonic. Consider part of the derivation shown below.
So, at your next jam session you can impress your friends with both your musical skills and Physics knowledge.
When morning rolls around an important element to starting the day on the right foot is a cup of coffee. But, did you know that every time you brew a cup of coffee you are essentially performing an extraction experiment? In this case, the extraction process is more specifically referred to as leaching. Leaching is a technique used to isolate a substance of interest from a solid by dissolving it in a liquid. Coffee grounds (inert solid) contain caffeine molecules (solute). The coffee grounds are submerged in water (solvent) to obtain the caffeine molecules. In water, the coffee grounds are insoluble while the caffeine molecules are soluble. Over time the caffeine molecules dissolve and diffuse into the water. When brewing coffee, a filter captures what remains of the coffee grounds but allows the caffeine and other molecules that have dissolved into the water to pass through. There was a study conducted to understand how time and water temperature affect the amount of caffeine obtained. The following charts were produced from data produced in the study for Coffea arabica beans[1]: Notice that the amount of dissolved caffeine in a solution increases over time and increasing temperature. Therefore, if completely disregarding taste, the longer and hotter you brew your coffee the more caffeine you will obtain. [1] Nhan, Pham Puoc, and Nguyen Tran Phu. "Effect of Time and Water Temperature on Caffeine Extraction from Coffee". Pakistan Journal of Nutrition. 11. 2. (2012): 100103. Web.
A popular fundraiser is to guess how much of a certain object are in a container. Often this object is sweet, sweet candy. The person with the closest guess wins the candyfilled container. Increase your odds of winning by making an educated guess! An educated guess is simply an estimate. Let's say there is a jar filled with jelly beans. How many jelly beans does the jar contain? First, estimate the size of the jar to determine its volume. You pull out the cell phone in your pocket that you guess is about six inches tall. Using it to estimate the dimensions of the jar, you determine the height is about oneandahalf cell phones and the diameter is one cell phone. Using the volume of the cylinder you calculate the volume of the jar to be about 254 cubic inches. Using the same logic you intuitively estimate that a jelly bean has a diameter of 0.5 inches and a length of 0.75 inches. Estimating the volume of a jelly bean using a cylinder you find the volume of a single jelly bean to be 0.15 cubic inches. Then it is a simple matter of dividing the volume of the jar by the volume of a single jelly bean to find that about 1,693 jelly beans could fit in the jar. But what about the gaps between the jelly beans? Great point! Various sources estimate the jar contains 20% air by volume [1][2]. Use proportions to calculate how many jelly beans would take up 80% of the jar by volume. How sweet is that? Give this technique a whirl at your next fundraiser. Who knows, your math wizardry may help you walk away with a jar full of jelly beans! [1] "How to win a guess the number of jelly beans contest". How Tutorial. WordPress. 18 April 2011. Web. 31 July 2016.
[2] "How to Win a Jellybean Guessing Contest". Cleverness: Getting Diggy with It. WordPress. 07 March 2007. Web. 31 July 2016. 
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