Question

I need help in this lab report, I am struggling, Please help, its urgent,
You can select data of your choice, like supposed data

PURPOSE: To calculate the acceleration due to gravity at the Earth's surface by measurement of the total distance an object falls and the time required to fall that distance. EQUIPMENT:1 tape measure or similar device 1 table tennis (Ping-Pong) ball 1 stopwatch EQUATIONS where v is the average velocity of the object, h is the height from which the object is dropped, and t is the time required for the object to fall. where veis the final velocity and v; is the initial velocity 2 Uf+0 vis zero since you dropped the ball = - 2 2 V 2v where v is the average velocity, v; is the initial velocity, and v is the final velocity of the object. f- Uf v at => a= _ where a is the acceleration of the object, vi is the initial velocity, v is the final velocity, and t is the time required for the object to fall PROCEDURE This lab is best performed with a partner, which allows one person to drop the ball and the other person to operate the stopwatch. Mark a convenient reference point from which the object will be dropped (example: the top of a door frame) 1. Measure the height of the reference mark (h) to the nearest 0.1 cm 2. Have the first partner hold the object at the mark 3. Have the second partner reset the timer to 0.00 4. Have one partner count to three, at which time the partners will simultaneously release the ball and start the timer 5. Have the second partner stop the timer when the ball strikes the ground and record the time (t) to the nearest 0.01s 6. Repeat the experiment (Trial 2) 7. Switch partners and repeat Step 1 through Step 7 (Trials 3-4) DATA Enter your data in the format below AVG SPEED FINAL TRIAL HEIGHT TIME (cm/s) SPEED (cm/s) ACCELERATION (cm/s2) V/t (cm) h NO. |(s) t v =h/t 2 v 1 2 3 1. Show the calculations for each of your trials. 1. DISCUSSION QUESTION: What is the accepted value of acceleration due to gravity, in units of cm/s2 (you must convert the textbook value 1 m 100cm), for comparison with your data?

Answer 1

Trial	Height h(cm)	Time t(s)	Avg Speed (m/s)V = h/t	Final speed (cm/s)Vf = 2v	Acceleration(cm/s2)a = Vf / t
1	2000	1.80	2000/1.8 = 1111.11	2x1111.11 = 2222.22	2222.22/1.8 = 1234.56
2	3000	2.80	3000/2.8=1071.42	2x1071.42= 2142.84	2142.84/2.8= 765.30
3	2000	2.10	2000/2.1= 952.38	952.38x2 = 1904.76	1904.76/2.1 = 907.03
4	3000	3.00	3000/2.6 =1153.85	1153.85x2= 2307.70	2307.7/2.6 = 887.57

Average acceleration(g) =

1234.56 765.3 907.03+ 887.57 3794.46/4 948.6cm/s2 a=

2)Standard g value = 980 cm/s²

If 10% error is accepted then

accepted range = 980 $\pm$ 10% of 980

                    Maximum value = 980 + 98= 1078 cm/s²

                     Minimum value = 980 -98 = 882 cm/s²

Our calculated value ( g = 948.6 cm/s² ) is within the accepted range.