SOLUTION: An arrow is fired straight up from the ground with an initial velocity of 192 feet per second. Its height, s(t), in feet at any time t is given by the function s(t)=-16t^2+192t. Fi

Algebra ->  Finance -> SOLUTION: An arrow is fired straight up from the ground with an initial velocity of 192 feet per second. Its height, s(t), in feet at any time t is given by the function s(t)=-16t^2+192t. Fi      Log On


   

Question 1095296: An arrow is fired straight up from the ground with an initial velocity of 192 feet per second. Its height, s(t), in feet at any time t is given by the function s(t)=-16t^2+192t. Find the interval of time for which the height of the arrow is greater than 252 feet.
Please show work.
Found 2 solutions by Alan3354, josmiceli:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
s(t)=-16t^2+192t = 252
Solve for t.
It's at or above 252 ft between the 2 times.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+s%28t%29+=+-16t%5E2+%2B+192t+
+-16t%5E2+%2B+192t+%3E+252+
+-4t%5E2+%2B+48t+-+63+%3E+0+
+t+=+%28+-b+%2B-+sqrt%28+b%5E2+-+4a%2Ac+%29%29+%2F+%28+2a+%29+
+a+=+-4+
+b+=+48+
+c+=+-63+
-----------------
+t+=+%28+-48+%2B-+sqrt%28+48%5E2+-+4%2A%28-4%29%2A%28-63%29+%29%29+%2F+%28+2%2A%28-4%29+%29+
+t+=+%28+-48+%2B-+sqrt%28+2304+-+1008+%29+%29+%2F+%28+-8+%29+
+t+=+%28+-48+%2B-+sqrt%28+1296+%29+%29+%2F+%28-8%29+
+t+=+%28+-48+%2B+36+%29+%2F+%28-8%29+
+t+=+12%2F8+
+t+=+3%2F2+
and also:
+t+=+%28+-48+-+36+%29+%2F+%28-8%29+
+t+=+84%2F8+
+t+=+10.5+
--------------------------
The interval of time for which +s%28t%29+%3E+252+ ft is
+1.5+%3C+t+%3C+10.5+ sec
---------------------
check:
here's the plot:
+graph%28++400%2C+400%2C+-3%2C+15%2C+-60%2C+600%2C+-16x%5E2+%2B+192x+%29+
Looks about right