Question

height	time1	time2	time3	time4	avg time
2 m	0.45 s	0.43 s	0.35 s	0.36 s	0.39 s
1.8 m	0.4 s	0.5 s	0.48 s	0.42 s	0.38 s
1.4 m	0.41 s	0.36 s	0.37 s	0.33 s	0.37 s
1.1 m	0.37 s	0.33 s	0.33 s	0.40 s	0.36 s

1) Using one of your heights and times from when the object was released to when the object hits the floor, calculate g. (This is an experimental value of g)

2) Using all of you height data points, figure out how to make a linear graph of height and time. Determine the slope of the graph. This is related to your second experimental value of g. Determine your second experimental value of g.

3) compare both your experimental values of g to the theoritcal value.

4) What is the percentage difference btween your experimetal values of g and the accepted value, 9.81m/s2 ?

5) How do you account fo this difference?

6) Does the theoritcal value for g fall within the final error range (the uncertainity) of your experimental value?

7) Explain with the primary limitation on your data and how it can be corrected. What factor in the experiment had the biggest effect on collecting valid data and how would you make the experiment more accurate?

please answer all questions and show work

Answer 1

1) When the object hits the ground the velocity will be equal to 0.

So initial velocoty=d/t=2/0.39=5.13 m/s

v=u-at

a=u/t=5.13 m/s/0.39=13.15 m/s²

b)Plot height along the y axis and time along the x axis

we will get a straight line. M

ark 2 points. Then find the slope.

For second value of g,

h=1.8 m and time t=0.38 s

final velocity=0 m/s

so v=u-gt

u=1.8/0.38=4.73 m/s

g=4.73 m/s/0.38=12.46 m/s²

c) Theorotical Value is 9.81 m/s²

Here we get in first case,g=13.15 m/s²

and second case,g=12.46 m/s²

4)V₁ = 9.81 and V₂ = 13.15

(( | V₁ - V₂ | / ((V₁ + V₂)/2) ) * 100

= ( | 9.81 - 13.15 | / ((9.81 + 13.15)/2) ) * 100
= ( | -3.34 | / (22.96/2) ) * 100
= ( 3.34 / 11.48 ) * 100
= 0.290941 * 100

= 29.0941% difference

V₁ = 9.81 and V₂ = 12.46

( | V₁ - V₂ | / ((V₁ + V₂)/2) ) * 100

= ( | 9.81 - 12.46 | / ((9.81 + 12.46)/2) ) * 100
= ( | -2.65 | / (22.27/2) ) * 100
= ( 2.65 / 11.135 ) * 100
= 0.237988 * 100 = 23.7988% difference

5) The accuracy is very low. Highly deviated from the original value