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January 25, 2015

January 25, 2015

Posted by **Alex** on Tuesday, February 12, 2008 at 9:15pm.

I first calculated the distance which is 160 m. Then I used the formula d=vit+1/2at^2. I got a quadratic -0.75t^2+22t-160. I got 13 or 16 as my time, but the answer says 15. Would it be correct to take the average of those times?

2) A stone is dropped from the roof of a high building. A second stone is dropped 1s later. How far apart are the stones when the second one has reached a speed of 23.0m/s?

I set this problem up by writing the given for stone 1 and stone 2.

Some equations I got were d(s1)= 4.905t^2 and ts2=ts1+1

I don't know how to do the problem.

- Physics -
**bobpursley**, Tuesday, February 12, 2008 at 10:01pm1. No, not the average.

vf=vi+at

0=22-1.5t

t=15seconds.

Your method rounded distance to 160m. If you had not rounded, you would have gotten 15 seconds....

2.find the time of the first to get to that speed.

vf=gt solve for t, adn the distance fell.

Now, subtract one from that t, and find out how far the second fell.

Subtract the distances.

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