In a movie clip the ball was released 7.42 m above the ground. In the same setti

Question

In a movie clip the ball was released 7.42 m above the ground. In the same setting the students dropped the ball 3 times and measured the times of fall with a stop watch. They recorded the following values: 1.33 s, 1.21 s and 1.24 s. From the average time of fall, calculate the experimental gravitational acceleration. Knowing that the uncertainty in the height measurement is H = 0.02 m and the average human response time is t = 0.2 s, estimate the uncertainty in your experimental g. Refer to the rules for error propagation presented in the introductory materials Basic Concepts of Error Analysis.

Explanation / Answer

h = 0.5gt^2
g = 2h/t^2

a. g1 = 2*7.42/1.33^2 = 8.3893 m/s/s
g2 = 2*7.42/1.21^2 = 10.1359 m/s/s
g3 = 2*7.42/1.24^2 = 9.6514 m/s/s
gav = 9.3922 m/s/s

b. dg/g = dh/h + 2dt/t
tav = (1.33+1.21+1.24)/3 = 1.26
h = 7.42
g = 9.3922

dg/9.3922 = 0.02/7.42 + 2*0.2/1.26
dg = 3.0069 m/s/s

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