Ball+on+incline-+Amir+raheem+and+Kevin+sarfani

Title of Lab: Ball on incline Lab

Researchers: Amir Raheem and Kevin Sarfani

Research Question : How does the time it takes for a Tennis ball to roll down on incline relate to the distance it takes

Hypothesis: The time is related to the distance: In this situation we assumed that initial velocity is zero: since the ball is moving in X direction we can assume that **mgx=** here Mass is can be ignored so therefore
 * d= vt+1/2at^2**
 * d=1/2at^2**
 * d/(1/2a)=t^2**
 * mgsinθ **
 * t^2=2d/9.81m/s^2* sintheta**

Procedure: materials: Tennis ball, an incline board, protractor, stop watch, measuring stick, chair, pencil
 * 1) Set an incline at angle 12 degrees
 * 2) Mark the measured distance with pencil
 * 3) Have a person roll down the ball and time it with a stop watch
 * 4) Repeat the steps 1-3

Data: This data represents the time taken for the ball to travel a given distance. The curve formed from this graph appears to form a square root curve. Taking the data and squaring the time gives us this data, which forms a straight line.
 * Distance (M) || Time (s) ||
 * 0 || 0 ||
 * 10 || 0.25 ||
 * 20 || 0.33 ||
 * 30 || 0.42 ||
 * 40 || 0.49 ||
 * 50 || 0.58 ||
 * 60 || 0.63 ||
 * 70 || 0.7 ||
 * 80 || 0.79 ||
 * 90 || 0.89 ||
 * 100 || 0.94 ||
 * 110 || 0.96 ||
 * 120 || 1.02 ||
 * 130 || 1.06 ||
 * 140 || 1.11 ||
 * Distance (M) || Time^2 (s) ||
 * 0 || 0 ||
 * 10 || 0.0625 ||
 * 20 || 0.1089 ||
 * 30 || 0.1764 ||
 * 40 || 0.2401 ||
 * 50 || 0.3364 ||
 * 60 || 0.3969 ||
 * 70 || 0.49 ||
 * 80 || 0.6241 ||
 * 90 || 0.7921 ||
 * 100 || 0.8836 ||
 * 110 || 0.9216 ||
 * 120 || 1.0404 ||
 * 130 || 1.1236 ||
 * 140 || 1.2321 ||

Data Analysis:

This graph is a Distance (meters) vs. Time (seconds) graph. This shows that as the distance gets longer, the time is takes to cover that distance gets longer as well. It is noticeable that this resembles a square root curve. We then consider squaring the data to get the next graph.

This graph is a Distance (meters) vs. Time^2 (seconds) graph. The graph appears to be much more straight and we then add the line of best fit.

The time squared vs displacement graph has a slope of 0.87 s^2/m, and using the equation g= 2/(slope*sin θ), we find the experimental g is equal to 11.06 m/s^2. According to the line of best fit, the y-intercept is -0.012 s^2.



slope: (1.23-0)/(140-0)= **.087** Conclusion: Our data supported our hypothesis, but the data slightly inaccurate but it can be fixed using more accurate timer for the ball instead of simply timing it by hand. With a theoretical value of .1 slope .087/.1 gives a 13% error.