Question

A projectile is fired straight up from ground level with an initial velocity of 112 ft/s. Its height, h, above the ground after t seconds is given by
h = –16t^2 + 112t. What is the interval of time during which the projectile's height exceeds 192 feet?
a. 3 < t < 4
b. t < 4
c. t > 4
d. 3 > t > 4

Answer 1

So you set h to 192, then set it up as an inequality
\[192ft <(112ft/s)t-(16ft/s^2)t^2\]
So the 112t - 16t^2 has to be greater than 192 ft. Then you come up with the polynomial

\[-(16ft/s^2)t^2+ (112ft/s)t-192ft > 0\]
to make arithmetic simpler, that you can reduce to
\[-(1ft/s^2)t^2+(7ft/s)t-(12ft)=0\]
of the form
\[at^2+bt+c>0\]

and solve for t using the quadratic formula - time will be greater than the smaller root, and less than the larger root

\[ t<\frac{-b-\sqrt{b^2-4ac}}{2a} \quad ; \quad t>\frac{-b+\sqrt{b^2-4ac}}{2a}\]
\[a=-1
\\ b=+7
\\ c=-12\]

Answer 2

so its D?