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by Quirkie on September 26th, 2007
A falling object's speed increases linearly with time.
It's average speed over the last 39m was 39/1.25 = 31.2m/s
It's change in speed is given by v = a t where a is about 10 m/s^2
It changed speed 12.5 m/s
It must have changed speed from 31.2-6.25 to 31.2+6.25 m/s finishing at 37.45 m/s
Which means it fell for 3.745 s, using v = a t again.
Using s = u t + (1/2) a t^2
s = 0 + 5 * 3.745^2
= 70.1m to three significant figures
This is assuming that it fell from stationary and that a=g=10m/s^2
Check: time to reach 70.1-39 m = sqrt(31.1/5) = 2.5 s
time to reach 70.1 m = sqrt(70.1/5) = 3.74 s
Difference = 1.26 s which is close enough because I rounded the distance figure.
Question: why thirty-nine? It's an odd number for a random question. Is this a real problem?
by WeeWillyWinky on September 25th, 2007
can you solve this with only the info provided?
Edit: nevermind i didnt think it through... gravity is constant so therefore if you figure out the average speed during those last 39 m then then the rate of acceleration you could in fact figure this some way or another i havent taken physics in years but it just boils down to algebra
When did Isaac Newton make the three laws of motion?
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You're reading A certain free-falling object requires 1.25 s to travel the last 39.m before it hits the ground. From what height above the ground did it fall?
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