A ball, and a feather dropped at the same height will hit the ground at the same time when placed in a vacuum space. This is because acceleration of motion is the same for both object, and in this experiment, we are going to calculate accelerated motion using inclined planes. In this experiment, we will see whether “distance traveled is directly proportional to the square of the time”, and “if speed of a falling object depend only on the height from which it falls”. This experiment will give us the idea of the variables involved in accelerated motion. Data Collection: In the experiment, we had a ramp, a jack, a can, a stopwatch, a 1 meter ruler, and a foot ruler. In part 1 of the experiment, we placed the jack 10cm high, placed one end of
…show more content…
At each height we placed the can at the 1 meter mark, and let go of the can while timing it until it hit the table. We repeated this step 6 times at each height. We took the average time of each height, then squared it, then divided the distance by the squared time, and divided the height by the length. Lastly we divide the distance by squared time by the height by length. For height 1, (2.44 + 1.85 + 1.78 + 2.11 + 2.44 + 2.31)s/6 = 2.16s, (2.16s)2 = 4.67, 1m/4.67s2 = .214m/s2, .1m/1m = .1, (.214m/s2)/.1 = 2.14m/s2. For height 2, (1.59+2.04+2.06+1.9+1.91+1.71)s/6 = 1.87s, (1.87s)2 = 3.5s2, 1m/3.5s2 = .286m/s2, .15m/1m = .15, (.286m/s2)/.15 = 1.90m/s2. For height 3, (1.58+1.39+1.52+1.39+1.45+1.45)s/6 = 1.46s, (1.46s)2 = 2.13s2. 1m/2.13s2 = .469m/s2, .2m/1m = .2, (.496m/s2/.2) = 2.35m/s2. For height 4, (1.2+1.32+1.26+1.26+1.19+1.26)s/6 = 1.25s, (1.25s)2 = 1.56s2, 1m/1.56s2 = .641m/s2, .25m/1m = .25, (.641m/s2)/.25 = 2.56m/s2.
Table 2A
L = 1m d=1.0m
Acceleration of the Can on the Incline Plane at Different Heights h(m) T1 (s) T2 (s) T3 (s) T4 (s) T5 (s) T6 (s) TAvg(s) T¬2 (s2) d/t2 h/L
.1 2.44 1.85 1.78 2.11 2.44 2.31 2.16 4.67 .214 .1
.15 1.59 2.04 2.06 1.9 1.91 1.71 1.87 3.5 .286 .15
.2 1.58 1.39 1.52 1.39 1.45 1.45 1.46 2.13 .469 .2
.25 1.2 1.32 1.26 1.26 1.19 1.26 1.25 1.56 .641
At each height we placed the can at the 1 meter mark, and let go of the can while timing it until it hit the table. We repeated this step 6 times at each height. We took the average time of each height, then squared it, then divided the distance by the squared time, and divided the height by the length. Lastly we divide the distance by squared time by the height by length. For height 1, (2.44 + 1.85 + 1.78 + 2.11 + 2.44 + 2.31)s/6 = 2.16s, (2.16s)2 = 4.67, 1m/4.67s2 = .214m/s2, .1m/1m = .1, (.214m/s2)/.1 = 2.14m/s2. For height 2, (1.59+2.04+2.06+1.9+1.91+1.71)s/6 = 1.87s, (1.87s)2 = 3.5s2, 1m/3.5s2 = .286m/s2, .15m/1m = .15, (.286m/s2)/.15 = 1.90m/s2. For height 3, (1.58+1.39+1.52+1.39+1.45+1.45)s/6 = 1.46s, (1.46s)2 = 2.13s2. 1m/2.13s2 = .469m/s2, .2m/1m = .2, (.496m/s2/.2) = 2.35m/s2. For height 4, (1.2+1.32+1.26+1.26+1.19+1.26)s/6 = 1.25s, (1.25s)2 = 1.56s2, 1m/1.56s2 = .641m/s2, .25m/1m = .25, (.641m/s2)/.25 = 2.56m/s2.
Table 2A
L = 1m d=1.0m
Acceleration of the Can on the Incline Plane at Different Heights h(m) T1 (s) T2 (s) T3 (s) T4 (s) T5 (s) T6 (s) TAvg(s) T¬2 (s2) d/t2 h/L
.1 2.44 1.85 1.78 2.11 2.44 2.31 2.16 4.67 .214 .1
.15 1.59 2.04 2.06 1.9 1.91 1.71 1.87 3.5 .286 .15
.2 1.58 1.39 1.52 1.39 1.45 1.45 1.46 2.13 .469 .2
.25 1.2 1.32 1.26 1.26 1.19 1.26 1.25 1.56 .641