Can anyone explain why C is wrong and correct it please, I have two attempts lef

Question

Can anyone explain why C is wrong and correct it please, I have two attempts left !!

You are a student in a science class that is using the following apparatus to determine the value of g. Two photogates are used. One photogate is located at the edge of a 1.00-m-high table and the second photogate is located on the floor directly below the first photogate. You are told to drop a marble through these gates, releasing it from rest at the same height as the table. The upper gate starts a timer as the ball passes through it. The second photogate stops the timer when the ball passes through its beam. Prove that the experimental magnitude of free-fall acceleration is given by g_exp = (2 delta y)/(delta t)^2, where delta y is the vertical distance between the photogates and delta t is the fall time. (Do this on paper. Your instructor may ask you to turn in this work.) For your setup, what value of 6t would you expect to measure, assuming g_exp is the standard value (9.81 m/s^2) During the experiment, a slight error is made. Instead of locating the first photogate even with the top of the table, your not-so-careful lab partner locates it 0.50 cm lower than the top of the table. However, she does manage to properly locate the second photogate on the floor directly below the first. What value of gexp will you and your partner determine What percentage difference does this represent from the standard value of g

Explanation / Answer

c)during the constant acceleration equation the speed of the marble,

vf2=vi2+2ax

vi=0 and a=g

So,

vf2=2gx

vf=sqrt(2gx)

Let v1 is the speed of the ball when it fallen 0.50 cm,v1=sqrt(2*9.8*0.005)=0.313 m/s

Let v2 is the speed of the ball when it fallen 0.50 m,v1=sqrt(2*9.8*0.5)=3.13 m/s

We can express time t as,t=v2-v1/g=3.13-0.313/9.8=0.2874 s

Distance that the marble moves=1-0.5-0.005=0.495 m

acceleration,g=2y/t2=2*0.495/(0.28742)=11.98 m/s2

Percenage difference=9.8-11.98/9.8=22.2%

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