SOLUTION: a man standing on a bridge drops a coin into a stream from a height of 112 feet. its distance above the stream after "t" seconds is given by h(t)-112ft-16t^2 (squared). find th

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Question 30500: a man standing on a bridge drops a coin into a stream from a height of 112 feet. its distance above the stream after "t" seconds is given by
h(t)-112ft-16t^2 (squared). find the coins average speed for teh first 2 seconds. what is the average speed of the coin between 2 seconds and the time the coin hits the water?
Answer by mbarugel(146) About Me  (Show Source):
You can put this solution on YOUR website!
The formula for the distance above the stream is
h%28t%29=112-16t%5E2
At t=0, the distance is 112 ft. After exactly 2 seconds pass (when t=2), the distance becomes 112-16%2A2%5E2+=+48+ft. This means that the coin has fell exactly 112 - 48 = 64 ft in 2 seconds. Therefore, the average speed is 64/2 = 32 ft/second.
The coin hits the water when the distance formula is zero. That is,
0=112-16t%5E2
112=16t%5E2
112%2F16=7=t%5E2
t=sqrt%287%29=+2.645...
Recall that at t=2, the coin was at a distance of 48 ft, while at t=2.645.. the distance wille be at a distance of zero. Therefore, it has come 48 ft in 0.645... seconds. So the average speed from t=2 to the time it touches the water is 48%2F%28sqrt%287%29-2%29+=+74.33 ft/second
I hope this helps!
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