SOLUTION: An object is thrown upward from the top of a 160 foot building with an initial velocity of 48 feet per second. The height h of the object after t seconds is given by the quadratic

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Question 480853: An object is thrown upward from the top of a 160 foot building with an initial velocity of 48 feet per second. The height h of the object after t seconds is given by the quadratic equation h= 16t2+48t+160. When will the object hit the ground?
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
h=0 when it reaches the terrace height on its way down
+h=+16t%5E2+%2B+48t+%2B+160
+0=+16t%5E2+%2B+48t+%2B+160
+0=+t%5E2+%2B+3t+%2B+10
solve using quadratic formula
t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a+=+1
b+=+3
c+=+10
t+=+%28-3+%2B-+sqrt%28+3%5E2-4%2A1%2A10+%29%29%2F%282%2A1%29+
t+=+%28-3+%2B-+sqrt%28+9+-+40+%29%29%2F2+
The square root is negative which is absurd.
The equation should be
+h+=+-16%5E2+%2B+48t+%2B+160
+0+=+-t%5E2+%2B+3t+%2B+10
a = -1,b = 3,c = 10
t+=+%28-3+%2B-+sqrt%28+3%5E2-4%2A%28-1%29%2A10+%29%29%2F%282%2A%28-1%29%29+
t+=+%28-3+%2B-+sqrt%28+9+%2B+40+%29%29%2F-2+
t+=+%28-3+%2B+7%29%2F-2 (ignore negative number)
t+=+%28-3+-7%29%2F-2
t = 5
The object will hit the ground in 5 sec
m.ananth@hotmail.ca