4. A body is dropped from rest and falls freely. Determine the position and spee

Question

4. A body is dropped from rest and falls freely. Determine the position and speed of the body after time: a) t = 1s; b) t = 2s; c) t = 3s; and d) t = 4s. (show your work) Hint: Choose the starting point as the origin. Initial velocity (at time t=0) is zero. To find the position use the kinematic equation for free-fall. To find the speed use equation for instantaneous velocity of a body undergoing constant acceleration (note: a = -g).

5. In lab #2 we used LINEST (Excel function) to determine the slope and uncertainties of the slope. Imagine that you are performing the following measurements: X Y 1 2.2 2 3.1 3 4.5 4 6.3 5 7.5 6 8.3 7 9.7 8 10.5 9 11.2 10 12.5

5.a. Using LINEST, calculate the slope and uncertainty of the slope for given data. Hint: Set your y column as your dependent variable and x column as your independent variable.

5.b. Calculate the mean value x , the standard deviation V y ,and standard deviation of the mean y of your ‘measured’ quantity X over N trials (number of runs). Xi (units) 0 .65 0.68 0.62 0.7 0.67 0.63 0.61 0.6 0.66 0.64

PLEASE SHOW WORK

Explanation / Answer

I am allowed to answer only 1 at a time. I will answer 1st question for you that is 1uestion 4

4.

Vi = 0m/s
a = -9.8m/s^2
a)
at t = 1 s
Vf = Vi + a*t
= 0 - 9.8*1
= -9.8 m/s
d = Vi*t + 0.5*a*t^2
= 0*1 + 0.5*(-9.8)*1^2
= - 4.9 m
b)
at t = 2 s
Vf = Vi + a*t
= 0 - 9.8*2
= -19.6 m/s
d = Vi*t + 0.5*a*t^2
= 0*2 + 0.5*(-9.8)*2^2
= - 19.6 m

c)
at t = 3 s
Vf = Vi + a*t
= 0 - 9.8*3
= -29.4 m/s
d = Vi*t + 0.5*a*t^2
= 0*3 + 0.5*(-9.8)*3^2
= - 44.1 m

d)
at t = 4 s
Vf = Vi + a*t
= 0 - 9.8*4
= -39.2 m/s
d = Vi*t + 0.5*a*t^2
= 0*1 + 0.5*(-9.8)*4^2
= - 78.4 m

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