SOLUTION: A stone is dropped from the top of a tall building and 1 sec later a second stone is thrown vertically down with a velocity of 60 ft/sec. How far below the top of the building will

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Question 895478: A stone is dropped from the top of a tall building and 1 sec later a second stone is thrown vertically down with a velocity of 60 ft/sec. How far below the top of the building will the second stone overtake the first?
Answer by thesvw(77) About Me  (Show Source):
You can put this solution on YOUR website!
Okay. First know The gravitational acceleration is 33ft/s (33 ft per second)
The first stone accelerates downward. The second stone has a initial velocity of 60ft/s. Note that it also accelerates.
The distance covered by both stones from the top should be equal when they meet. Lets get it as 'x'.
Lets think about the time taken by each stone. 2nd stone was thrown after 1 second. That means 1st stone has a bonus 1 second to move.
Time taken by 1st stone = %28t%2B1%29
" " " 2nd stone = t
Now hope you know this equation..
s+=+ut+%2B+%281%2F2%29%2Aat%5E2
Apply the know data there ...
For the first stone x+=+%281%2F2%29%2Ag%2A%28t%2B1%29%5E2
2nd stone x+=+60t+%2B+%281%2F2%29%2Ag%2At%5E2
g is the gravitational acceleration.
Both equations are equal. Solve them for t
Well .. I got the value of t as..
t+=+g%2F2%2A%2860-g%29
Now apply the value and find x from the above equations we wrote earlier.
I got g+=+33+ft%2Fsec
I got the answer x+=+48.82ft The distance from the top.