Falling object problem Calculation I drop a rock from the top of a building. A f
ID: 2106082 • Letter: F
Question
Falling object problem Calculation I drop a rock from the top of a building. A few moments later I take two speed measurements (with one of those police radar guns), and find that the second speed measurement is 2x as big as the first. If the time between those measurements is 0.5 seconds, how far from the top of the building was it when I took the first measurement? Hint: Draw yourself a graph of velocity vs. time for a falling object that starts from rest, find two points that are 0.5 seconds apart and one has twice the speed of the other, and you'll be able to figure out how long it's been since it started falling. Follow-up I measure the distance that my object traveled at the following times: The time of the first speed measurement, the time of the second speed measurement, and 0.5 seconds after the second speed measurement. What is the ratio y3/y1, where y3 is the third distance measurement and y1 is the first? Note that the distances are all measured with respect to the roof that I dropped the object from, so each y measurement is how far it has fallen since being dropped, not how far it has fallen since the previous measurement.Explanation / Answer
v1 = v = gt1
v2 = 2 v1 = gt2
v2 -v 1= g (T2-T1)
v= 10 X 0.5
= 5 ms^-1
t1 = v/g
= 0.5 s
y1 = 1/2 g t ^2 = 1.25 m
t2 = t1 + 0.5
= 1 s
y3 = 1/2 g t^2
= 0.5 X10 1
= 5 m
y3-y1 = 3.75 m
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