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. 
:
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 = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this problem a=-16, b=14; c=-3
{{{t = (-14 +- sqrt(14^2-4*-16*-3 ))/(2*-16) }}}
:
{{{t = (-14 +- sqrt(196 - 192))/(-32) }}}
:
{{{t = (-14 +- sqrt(4))/(-32) }}}
Two solutions:
{{{t = (-14 + 2)/(-32)}}}
{{{t = (-12)/(-32)}}}
t = .375 sec; (passes 103 ft on the way up)
and
{{{t = (-14 - 2)/(-32)}}}
{{{t = (-16)/(-32)}}}
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?