Ariel+and+Allison


 * Lab: Bombs Away**

**Group Members:** Allison Irwin and Ariel Katz **Class:** Period 2 **Date Completed:** 9/27/10 **Date Due:** 9/28/10

**Objective:** What is acceleration due to gravity? **Materials:** spark timer, spark tape, mass, tape measure

**Hypothesis:** When gravity equals 9.8g, acceleration will equal 10m/s2 because the change in velocity relative to time will equal 10m/s. what is 9.8g?

**Procedure:**
 * 1) Attach spark timer to cabinet with clamp.
 * 2) Feed about 1m piece of spark tape through the timer.
 * 3) Attach a 100g weight to the spark tape with tape.
 * 4) Switch spark timer to 60hertz setting.
 * 5) Turn on spark timer and drop the weight at the same time.
 * 6) Mark all points on spark tape.
 * 7) Measure distance of each point.
 * 8) Create a Distance-Time table with these points on Excel.
 * 9) Perform calculations to find acceleration due to gravity.
 * 10) Find percent error.

**Data:** **need units in headings of a spreadsheet.** **Calculations:**

Calculations: Y=405.01x2-0.7452x Y=Ax2+Bx D=At2 + Bt D=Vit+1/2at2 A=1/2a 405.01=1/2a
 * a= 8.100 m/s2**

this is a large percent error

**Discussion Questions:**

Conclusion?
 * 1) Does the shape of your graph agree with the expected graph? Why or why not?
 * Yes, it does. Falling objects' speeds increase as they fall. The position time graph is consistent with a velocity time graph of increasing speed. Also, from the graph, velocity can be calculated by observing the change in position divided by the change in time. If that is done for the graph, we will see that the velocity is increasing as the object continues to fall. The slope at the beginning of the graph is not as steep as the slope at the end of the graph. The steeper the slope is, the greater the velocity is.
 * We can also look at the R^2 value, which is 0.9978. The R2 value is very close to one, or 100%, which implies accuracy.
 * 1) How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.)
 * The class average was 926.39 cm/s2. Our acceleration was 810.0 cm/s2. We can find the percent difference by calculating the absolute vale of the difference between the class's value and our value: |926.39-810.0| and dividing the result by the class average, and get .1256. Then, we multiply that value by 100, to find that the percent difference equals 12.56%.
 * 1) Did the object accelerate uniformly? How do you know?
 * In this lab the acceleration was due to gravity. Because gravity is a constant force on a free-falling object, the object accelerated uniformly as it fell. The acceleration constant due to gravity is g= 9.81m/s^2.
 * 1) What should the velocity-time graph of this object look like?
 * The velocity time graph of this object should be a diagonal line with a positive slope. The object's velocity should start at zero and end with a large velocity because it is getting faster. Because acceleration due to gravity is 9.81m/s^2, the slope of the line should be 9.81. good
 * 1) Write down the expected equation of the line from this v-t graph (use specific information from your x-t graph).
 * y = 810.02x
 * 1) What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?
 * Acceleration due to gravity would be higher if someone or something (ex. wind) was pushing down on the object as it fell.
 * Acceleration due to gravity would be lower than it should be if something was keeping the object from freely falling. In our lab, if the spark tape did not flow freely through the timer, acceleration due to gravity would have been lowered. It also would have ben lowered if it hit something (ex. cabinet, counter, etc.) while it attempted to fall.