Purpose: Measuring ‘g’ by dropping steel ball and Ping-Pong ball from the same height and record the time for each dropping. Then, using the equation “h=1/2g*t^2 to calculate ‘g’. For precise, we still need to find ∆g, therefore, the final data about g=gave+∆g.
Method: Before we start to do the experiment, we need to find a position which can except the wind and flat ground. After that, two people as a group, one is a golfer, and another is a dropper. First, they measured use tape measure to measuring the distance between dropping place and ground, and at lease twice (each student do once). For reduce the uncertainty, usually choose more than 5 meters high place. When starting the experiment, one student as dropper stay at the high place, …show more content…
Explain. (Note: Timing technique may help, but that’s a different matter!!)
Yes, a more precise stop watch will reduce the uncertainty. For example, compare with the αt, which is “standard deviation of the mean” between the first one (0.038s) and the second one (0.040s) which use the same stop watch. Obviously, the second one is a little greater than the first. Also, the uncertainty average time in same line smaller as well (0.055s<0.067s). As a result, using more precise stop watch will decrease the value of the …show more content…
Based on this calculation, would timing more drops help in reducing the uncertainty of the average times? In this experiment would it be feasible to take the optimal number of measurements?
In my works, there are at least 5 times of each ball to measurement need. With the calculation, If student timing more drops, it will help in decreasing the uncertainly of the average time. So, 5 times of dropping are enough for this experiment.
Q9: Were the times given by the preferred method for the first ball and the second ball significantly different? (Include your actual calculated values in your explanation.)
No, there is not too much different. Through mine calculate, when the student A as a dropper, the average time is 1.026s, however, when A as a golfer, the average time is 0.902s. When they change to the second ball (Ping-Pong ball), A is a dropper and a timer as well, the average time is 0.968. Therefore, it has not significantly different (1.026-0.968=0.058s).
Q10: Would a more precise device to measure h reduce the uncertainty in g or not? Explain.
Yes, if decrease the uncertainty of the h, the g will reduce the uncertainty as well. Since ∆g = (∆h/h +∆t/t)*(2h/t^2), when the value of ∆h decreasing, ∆g will reduce as
Method: Before we start to do the experiment, we need to find a position which can except the wind and flat ground. After that, two people as a group, one is a golfer, and another is a dropper. First, they measured use tape measure to measuring the distance between dropping place and ground, and at lease twice (each student do once). For reduce the uncertainty, usually choose more than 5 meters high place. When starting the experiment, one student as dropper stay at the high place, …show more content…
Explain. (Note: Timing technique may help, but that’s a different matter!!)
Yes, a more precise stop watch will reduce the uncertainty. For example, compare with the αt, which is “standard deviation of the mean” between the first one (0.038s) and the second one (0.040s) which use the same stop watch. Obviously, the second one is a little greater than the first. Also, the uncertainty average time in same line smaller as well (0.055s<0.067s). As a result, using more precise stop watch will decrease the value of the …show more content…
Based on this calculation, would timing more drops help in reducing the uncertainty of the average times? In this experiment would it be feasible to take the optimal number of measurements?
In my works, there are at least 5 times of each ball to measurement need. With the calculation, If student timing more drops, it will help in decreasing the uncertainly of the average time. So, 5 times of dropping are enough for this experiment.
Q9: Were the times given by the preferred method for the first ball and the second ball significantly different? (Include your actual calculated values in your explanation.)
No, there is not too much different. Through mine calculate, when the student A as a dropper, the average time is 1.026s, however, when A as a golfer, the average time is 0.902s. When they change to the second ball (Ping-Pong ball), A is a dropper and a timer as well, the average time is 0.968. Therefore, it has not significantly different (1.026-0.968=0.058s).
Q10: Would a more precise device to measure h reduce the uncertainty in g or not? Explain.
Yes, if decrease the uncertainty of the h, the g will reduce the uncertainty as well. Since ∆g = (∆h/h +∆t/t)*(2h/t^2), when the value of ∆h decreasing, ∆g will reduce as