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
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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) (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:
In this problem a=-16, b=14; c=-3
:
:
Two solutions:
t = .375 sec; (passes 103 ft on the way up)
and
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?
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