Ball+on+an+Incline+-+Holly+and+Corey

Title of Lab: Ball on an Incline

Researchers: Holly Therrell and Corey Keyser

Research Question: How does the time it takes for a tennis ball to roll down an incline relate to the distance that it travels?

Research: Based on the idea that the acceleration of gravity going down is 9.81 m/s^2 and acceleration is a vector, we weer able to surmise that the the acceleration for an incline can be found based on the properties of triangles using the formula a=9.81(sin(theta)), where a is acceleration along incline, and theta is angle of inclination. Using this knowledge we are able to plug this into the kinematics equation (delta)x=Vit+.5at^2.This provides us with the equation (delta)x=.5(gsin(theta))t^2,ViT is thus equal to zero because Vi is defined as zero in the procedure. This process thus gives us an equation that relates distance traveled to speed.

Hypothesis: The time it takes for a tennis ball to roll down an incline is in quadratic,direct relation to the distance traveled as depicted by the equation t^2=2(delta)x/(gsin(theta)). Where (delta)x is distance traveled, theta is angle of inclination, and t is time.

Equipment: stopwatch, wood board, tennis ball, meter stick, protractor

Procedure: 1. Set up board to having an angle of inclination of 15 degrees. 2. Set textbook at bottom as a ball stop. 3. Measure distance from ball stop up the board. 4. One person both drops the ball and operates the stopwatch in order to have the most accurate start times. 5. Record time it takes from beginning position till it hits textbook stop. 6. Repeat 3 times at each distance. 7. Change distance by 10 cm, starting at 100 cm and going till 0. Controls: board, tennis ball, stop watch, dropper and stop watch recorder, angle of inclination

Data:
 * ΔX || Time (trial 1) || Time (trial 2) || Time (trial 3) || Time (average) ||
 * 100cm || 0.9s || 1.1s || 0.9s || 1.0s ||
 * 90cm || 0.9s || 0.9s || 0.9s || .9s ||
 * 80cm || 0.8s || 0.8s || 0.8s || .8s ||
 * 70cm || 0.7s || 0.7s || 0.7s || .7s ||
 * 60cm || 0.7s || 0.7s || 0.7s || .7s ||
 * 50cm || 0.6s || 0.6s || 0.6s || .6s ||
 * 40cm || 0.5s || 0.5s || 0.5s || .5s ||
 * 30cm || 0.4s || 0.4s || 0.5s || .4s ||
 * 20cm || 0.3s || 0.4s || 0.3s || .3s ||
 * 10cm || 0.2s || 0.2s || 0.1s || .2s ||
 * 0cm || 0.0s || 0.0s || 0.0s || 0 ||

Data Analysis: In the graph of our data below, the line drawn is our line of best fit. Using the two points circled, our slope (2/gsin(theta)) was found to be 0.01 s^2/cm, or 1 s^2/m, which means our experimental g was 7.7 m/s^2. The y-intercept of our data is (0cm,0s^2), which means it took 0s to travel 0cm, which makes sense because it should take no time to travel no distance. However, the y-intercept of the line of best fit is (0cm,-0.09s^2), which doesn't make sense because it shouldn't take negative time to travel no distance and when you square something it can never be negative.

(can't figure out how to flip graph)

Conclusion: Our experimental g ended up being 7.7 m/s^2 compared to the 9.81 m/s^2 we should have gotten. This discrepancy gave us a large error of 21%. Even though reasoning behind the equation seemed sound, the data doesn't follow the intended graph very well. Though error was highly likely due to our imprecise stopwatch and human inability to stop the time right when the ball finished rolling, the data also leads us to believe that our hypothesis is incorrect. Perhaps some other factor is involved that keeps the acceleration from reaching 9.81 m/s^2. We could improve the experiment by recording the time using light gates that the ball would pass through, but we would need more physics knowledge to design an experiment that reveals why acceleration on an incline is less than free fall acceleration.