SOLUTION: A ball is thrown straight up with an initial velocity of 80 feet per second from the top of a building 23 feet tall. It's height h (in feet) above the ground t seconds after the b

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Question 517795: A ball is thrown straight up with an initial velocity of 80 feet per second from the top of a building 23 feet tall. It's height h (in feet) above the ground t seconds after the ball is thrown is given by h = -16t^2 + 80t + 23. When will the ball be 119 feet above the ground for the first time?
*** ^2 stands for squared or to the second power ***
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The ball will probably be 119' above the ground twice:
once going up and again coming down.
Just pick the smallest value for t and that will
be going up.
+h+=+-16t%5E2+%2B+80t+%2B+23+
+119+=+-16t%5E2+%2B+80t+%2B+23+
+-16t%5E2+%2B+80t+%2B+23+-+119+=+0+
+-16t%5E2+%2B+80t+-+96+=+0+
+-t%5E2+%2B+5t+-+6+=+0+
Use quadratic formula
+t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a=+-1+
+b+=+5+
+c+=+-6+
+t+=+%28-5+%2B-+sqrt%28+5%5E2+-+4%2A%28-1%29%2A%28-6%29+%29%29%2F%282%2A%28-1%29%29+
+t+=+%28-5+%2B-+sqrt%28+25+-+24+%29%29%2F%28-2%29+
+t+=+%28+-5+%2B+1+%29+%2F+%28-2%29+
+t+=+2+
and, also,
+t+=+%28+-5+-+1%29+%2F+%28-2%29+
+t+=+3+
2 sec from when the ball is thrown, the
height will be 119' above ground for the first time.
check:
+119+=+-16t%5E2+%2B+80t+%2B+23+
+119+=-16%2A4+%2B+80%2A2+%2B+23+
+119+=+-64+%2B+160+%2B+23+
+119+=+96+%2B+23+
+119+=+119+
OK
Here's a plot:
+graph%28+400%2C+400%2C+-1%2C+6%2C+-10%2C+130%2C+-16x%5E2+%2B+80x+%2B+23+%29+