Basketball in the Snow
Basketball is mostly an indoor sport, there's no doubt about that. However, for this blog entry, I decided to venture out into the sub-zero temperatures for a fun game of snowy hoops. I decided to put my shooting skills to the test and analyze the components of my shot using physics. It didn't take long before the wind chill got to me and I learned why basketball games generally are not played outdoors in the dead of winter!
Below is the video I took of my shot attempt:
Firstly, we need to determine the known components of the shot.
The height of the player is about 6 feet, 1 inch. (1.85 m.)
The height of the basketball hoop is 10 feet. (3.05 m.)
The distance from the player to the basket is 12 feet. (3.66 m.)
The time it takes for the basketball to reach the net is 1.46 seconds (timed using a stopwatch)
Since this is a projectile motion problem, we need to break up our components into the x and y direction. Looking at our given components, it appears that we are still missing the acceleration and velocity of the ball.
The acceleration we can find fairly easily. In the y direction, the ball is being pulled back down to earth by the force of gravity, which is roughly -9.8 m/s^2. On the other hand, in the x direction, gravity has no effect on the ball so the acceleration is simply 0.
The velocity of the ball is a bit more tricky. We know that the velocity in the x direction is equal to Vi*cos(θ) and the velocity in the y direction is Vi*sin(θ), however we know neither the angle or the initial velocity of the ball.
Using a screenshot from my video, we can draw a picture of the problem.
Note: the 1.20 m was found by subtracting the height of the player from the height of the basketball hoop
Since this is a real photo of our problem (in contrast to the usual drawing), we can insert a protractor to get an estimate of the angle.
From the protractor, the angle appears to be about 74 degrees.
We can use these measurements to find the initial velocity by using the equation ∆x = 1/2at^2 + Vit
By plugging our values into the equation, we get:
3.66 = 1/2(0)(1.46)^2+(Vicos74)(1.46)
3.66 = (Vicos74)(1.46)
2.51 = Vicos74
9.09 = Vi
Vi = 9.09 m/s
This gives us an x velocity of 9.09*cos74 =2.51 m/s and a y velocity of 9.09*sin74 = 8.74 m/s
To verify our results, we can use LoggerPro to graph the path of the ball and determine an accurate velocity in both directions.
X velocity:
From LoggerPro, the slope for the best fit line of velocity is about 2.23 m/s. This is 0.28 m/s less than my calculations.
Y velocity:
The slope of the best fit line shows a slope of 7.44 m/s. This is 1.30 m/s less than the y velocity I calculated.
These slight errors probably came from the fact that the angle was an estimation. Using the inserted protractor, it was difficult to determine an exact angle of the release of the ball. However, the calculated initial velocity of 9.09 m/s seems to be reasonably accurate given that my results were not off by a wide margin.
Below is the video I took of my shot attempt:
Firstly, we need to determine the known components of the shot.
The height of the player is about 6 feet, 1 inch. (1.85 m.)
The height of the basketball hoop is 10 feet. (3.05 m.)
The distance from the player to the basket is 12 feet. (3.66 m.)
The time it takes for the basketball to reach the net is 1.46 seconds (timed using a stopwatch)
Since this is a projectile motion problem, we need to break up our components into the x and y direction. Looking at our given components, it appears that we are still missing the acceleration and velocity of the ball.
The acceleration we can find fairly easily. In the y direction, the ball is being pulled back down to earth by the force of gravity, which is roughly -9.8 m/s^2. On the other hand, in the x direction, gravity has no effect on the ball so the acceleration is simply 0.
The velocity of the ball is a bit more tricky. We know that the velocity in the x direction is equal to Vi*cos(θ) and the velocity in the y direction is Vi*sin(θ), however we know neither the angle or the initial velocity of the ball.
Using a screenshot from my video, we can draw a picture of the problem.
Note: the 1.20 m was found by subtracting the height of the player from the height of the basketball hoop
Since this is a real photo of our problem (in contrast to the usual drawing), we can insert a protractor to get an estimate of the angle.
From the protractor, the angle appears to be about 74 degrees.
We can use these measurements to find the initial velocity by using the equation ∆x = 1/2at^2 + Vit
By plugging our values into the equation, we get:
3.66 = 1/2(0)(1.46)^2+(Vicos74)(1.46)
3.66 = (Vicos74)(1.46)
2.51 = Vicos74
9.09 = Vi
Vi = 9.09 m/s
This gives us an x velocity of 9.09*cos74 =2.51 m/s and a y velocity of 9.09*sin74 = 8.74 m/s
To verify our results, we can use LoggerPro to graph the path of the ball and determine an accurate velocity in both directions.
X velocity:
Y velocity:
These slight errors probably came from the fact that the angle was an estimation. Using the inserted protractor, it was difficult to determine an exact angle of the release of the ball. However, the calculated initial velocity of 9.09 m/s seems to be reasonably accurate given that my results were not off by a wide margin.
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