Question

Y | t^2
40 | 2.42
35. | 2.59
30 | 2.77
25 | 2.98
20 | 3.17
15 | 3.34
10 | 3.57
Plot a graph t^2 vs y
Find the slope s
The intercept on t^2 axis,c
Derive the relationship between t^2 and y and relate S and c with the physical constants of the set up
Explain how this set up can be used to determine the value of g, limitations if any

Answer 1

slope = (3.57 - 2.42 ) / (10 - 40) = 1.15 / -30 = - 0.0383 = -0.04 to 2 figures
check to see if constant (can not graph here)
s = (3.17 - 2.77) / ( 20 - 30) = 0.4 / -10 = -0.04
t^2 = -0.04 y + b
find b
if y = 20 , t^2 = 3.34
3.34 = -.04*20 + b
b = 3.34 + 0.8 = 4.14
so
t^2 = -.04 y + 4.14
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well I do not know your setup but if it is a falling mass m at height h starting at Hi
then
a = dv/dt = -g
v = Vi - g t
if Vi = 0 then v = - g t
h = Hi - (1/2) g t^2
(g/2) t ^2 = (Hi-h) = -h + Hi