OpenStudy (anonymous):

when a ball is thrown straight down from the top of a building, with initial velocity 30ft/sec, the distance in feet from the release point at time t seconds is given by s(t) = 16t^2 + 30t. If the release point is 300 feet above the ground, what is the velocity of the ball at the time it hits the ground? a) 142 ft/sec b) 145 ft/sec c) 149 ft/sec d) 153 ft/sec

OpenStudy (anonymous):

$s(t) =1 6t^2 + 30t\\ v(t) = s'(t) = 32t + 30$ First thing's first: solve where s = 300. $s(t) =300\\ 16t^2 + 30t - 300 = 0\\ t = \frac{-30\pm\sqrt{30^2 - 4(16)(-300)}}{2(16)}\\ \ = \frac{-30\pm\sqrt{900 + 19200}}{32}\\ \ = \frac{-15+5\sqrt{201}}{16}$ (we throw away the extraneous negative value) Now, solve for the velocity. $v(t)\simeq111.774$

OpenStudy (anonymous):

Thanks for explaining that but that isn't one of the answer choices....

OpenStudy (anonymous):

Woops - it's A. I forgot to add 30 when computing v :-)

OpenStudy (anonymous):

Oh, thanks so muchh!