SOLUTION: A ball is thrown upward from a 100 foot tall building with an initial velocity of 14 feet per second. Its height s(t) in feet is given by the function s(t)= -16t^2+14t+100. Find

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A ball is thrown upward from a 100 foot tall building with an initial velocity of 14 feet per second. Its height s(t) in feet is given by the function s(t)= -16t^2+14t+100. Find       Log On


   

Question 150467: A ball is thrown upward from a 100 foot tall building with an initial velocity of 14 feet per second. Its height s(t) in feet is given by the function s(t)= -16t^2+14t+100. Find the interval of time for which the height of the ball is greater than 103 feet.
I don't even know where to start.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A ball is thrown upward from a 100 foot tall building with an initial velocity of 14 feet per second. Its height s(t) in feet is given by the function s(t)= -16t^2+14t+100. Find the interval of time for which the height of the ball is greater than 103 feet.
:
Write the equation for s = 103 ft
-16t^2 + 14t + 100 = 103
:
-16t^2 + 14t + 100 - 103 = 0
;
-16t^2 + 14t - 3 = 0
:
Use the quadratic formula to find t:
t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
In this problem a=-16, b=14; c=-3
t+=+%28-14+%2B-+sqrt%2814%5E2-4%2A-16%2A-3+%29%29%2F%282%2A-16%29+
:
t+=+%28-14+%2B-+sqrt%28196+-+192%29%29%2F%28-32%29+
:
t+=+%28-14+%2B-+sqrt%284%29%29%2F%28-32%29+
Two solutions:
t+=+%28-14+%2B+2%29%2F%28-32%29
t+=+%28-12%29%2F%28-32%29
t = .375 sec; (passes 103 ft on the way up)
and
t+=+%28-14+-+2%29%2F%28-32%29
t+=+%28-16%29%2F%28-32%29
t = .5 sec; (passes 103 ft on the way down)
:
So the interval of time would be .5 - .375 = .125 sec above 103 ft
;
How about this? Does it make sense to you now?